#Logarithm

203 messages · Page 1 of 1 (latest)

stone sparrowBOT
sharp bluff
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$\log_{a}\left(b\right)\cdot\frac{1}{\log_{a}\left(b\right)}$ is literally just one. thats why it cannot equal zero and why $\log_{20}\left(2\right)\cdot\frac{1}{\log_{20}\left(2\right)}$ is equal to one

daring ravineBOT
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Judgemental Snail

655194646523281428.png
surreal sierra
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No bro

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That's the log property it should be zero

sharp bluff
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$a\cdot\frac{1}{a}\to\frac{a}{a}\to1$

daring ravineBOT
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Judgemental Snail

655194646523281428.png
sharp bluff
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a=log_a(b) in this case

sharp bluff
surreal sierra
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1/log_b(a) can be written as log_b(a)^-1

sharp bluff
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uh huh.. x*x^-1=1

surreal sierra
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Bro have u studied log?

sharp bluff
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yes, what are you getting at?

surreal sierra
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Log_b(a)^n=?

sharp bluff
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are you referring to the power rule?

surreal sierra
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Yes

sharp bluff
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thats not how the power rule works

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log_a(b^n)=n(log_a(b))

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only works if the exponent is inside the parenthesis

surreal sierra
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Log_b(a)²= 2log_b(a) ryt?

sharp bluff
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nope

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Log_b(a²)= 2log_b(a)

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exponent gotta be inside the log

surreal sierra
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Oh sorry dis is what i meant

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While writing in paper we don't use parenthesis so got confused

sharp bluff
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oh you should use parenthesis more as to avoid situations like this one

surreal sierra
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So my question is log_b(a^-1) is - log_b(a) ryt?

sharp bluff
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yes, but not in this context

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1/log_a(b)=log_a(b)^-1

surreal sierra
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Den y did wolfram alpha gave wrong answer

sharp bluff
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it isnt the wrong answer, $\log_{a}\left(b\right)\cdot\frac{1}{\log_{a}\left(b\right)}$ can never equal 0

surreal sierra
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That's the log property dude

daring ravineBOT
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Judgemental Snail

655194646523281428.png
surreal sierra
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, w solve log_b(a) *1/log_b(a)

daring ravineBOT
surreal sierra
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See it is 0

sharp bluff
surreal sierra
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Rhs?

sharp bluff
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it assumed you mean 0

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you were not specific enough

surreal sierra
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U try den

sharp bluff
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, w simplify log_b(a) *1/log_b(a)

daring ravineBOT
surreal sierra
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, w simplify log_b(a) * log_b(a^-1)

daring ravineBOT
sharp bluff
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like i said, the ^-1 wil go outside the parenthesis

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,w simplify log_b(a) * log_b(a)^-1

daring ravineBOT
surreal sierra
sharp bluff
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power rule doesnt apply in this no

surreal sierra
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Just a min

surreal sierra
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, w simplify ln(x)+ln(1/x)

daring ravineBOT
surreal sierra
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, w solve ln(x)+ln(1/x)

daring ravineBOT
surreal sierra
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, w solve ln(x) + ln(1/x)

daring ravineBOT
surreal sierra
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, w simplify ln(2)*1/ln(2)

daring ravineBOT
surreal sierra
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, w solve ln(2)*1/ln(2)

daring ravineBOT
surreal sierra
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, w solve ln(2)*ln(1/2)

daring ravineBOT
surreal sierra
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, w simplify ln(2)*ln(1/2)

daring ravineBOT
surreal sierra
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, w solve ln(a+(a²+1)^1/2) +ln(1/a+(a²+1)^1/2)

daring ravineBOT
surreal sierra
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@buoyant quest

sharp bluff
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dont ping random users for help

surreal sierra
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He helped me in calculus

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N u weren't answering so

sharp bluff
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unless he told you that you can ping, dont

surreal sierra
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Ok

sharp bluff
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anywas is this still pertaining to this

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?

surreal sierra
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Bro sorry i misunderstood it should be natural log not common log

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Ln(x) *ln(1/x) this should give zero

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U there?

sharp bluff
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yes srry im helping you and another at the same time

surreal sierra
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Ohh

sharp bluff
surreal sierra
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Bro its the same question

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There i took some base

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But wen i rechecked my property found that it was natural log with base e

sharp bluff
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im not sure where you are deriving these equations from

surreal sierra
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Its properties of log

sharp bluff
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you might just be looking too much into this

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$\log_{a}\left(b\right)\cdot\frac{1}{\log_{a}\left(b\right)}=\frac{\log_{a}\left(b\right)}{\log_{a}\left(b\right)}=1$

daring ravineBOT
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Judgemental Snail

655194646523281428.png
surreal sierra
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Sorry log shouldn't be in denominator

sharp bluff
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why? this was your original question

surreal sierra
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Typing mistake

sharp bluff
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ohh... you prob could have provided that screen shot much sooner lol

surreal sierra
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I was searching from before bro

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In my notebook too i noted wrongly

sharp bluff
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log(x)+log(1/x)=log(x)+log(x^-1)=log(x)-log(x)=0

surreal sierra
sharp bluff
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all good

surreal sierra
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Now try in texit once

sharp bluff
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i didnt know that this was the problem you were refering to

surreal sierra
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It is still not giving 0

sharp bluff
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,w simplify log(x)+log(1/x)

daring ravineBOT
surreal sierra
sharp bluff
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it applies for all bases

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or at least most, there are a few undefined bases that i do not wish to look at

surreal sierra
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Y is not giving 0?

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, w solve log(x)+log(1/x)

daring ravineBOT
surreal sierra
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It again assumed my question to be zero?

sharp bluff
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ok what is your actual question, is it why this rule exists?

surreal sierra
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, w solve log(10)+log(1/10)

daring ravineBOT
sharp bluff
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im genuinely confused about what you are asking

surreal sierra
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Bro wt im asking is the general property and the inputs wen i put in that property is not matching

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But why is that so

sharp bluff
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what inputs are you putting in?

surreal sierra
surreal sierra
sharp bluff
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that matches the given rule

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that also matches the given rule

surreal sierra
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The graph i didn't understand can u pliz explain

surreal sierra
sharp bluff
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for all x inputs greater than 0, y=0

surreal sierra
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Here rhs didn't assume my question to be 0?

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, w solve ln(x)*1/ln(x)

daring ravineBOT
surreal sierra
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, w solve ln(x)+ln(1/x)

sharp bluff
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it should be "+"

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not *

daring ravineBOT
surreal sierra
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, w solve ln(x)+ln(1/x)

daring ravineBOT
sharp bluff
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that x>0 just means it holds true for all x values above 0

surreal sierra
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, w solve ln(x)+1/ln(x)

daring ravineBOT
surreal sierra
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Bro pliz clear this confusion

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, w solve Log_b(a)+log_b(1/a)

sharp bluff
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idk what you are confused about, i already addressed what all these results mean and even provided you a proof for the rule.

daring ravineBOT
surreal sierra
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, w solve Log_b(a)+1/log_b(a)

daring ravineBOT
sharp bluff
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ok, lets address this, what are you confused about?!? all these wolfram calculations are not gonna help. i need to know what exactly you are confused about so i can help you resolve that confusion.

surreal sierra
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It's basically related to wolfram result... Before u said in 1 question it assumed my question to be 0..... And even now it is giving as 0... So are these two zeroes different?

sharp bluff
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here, lets start over

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what is your question about the following rule: log(x)+log(1/x)=0?

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ignoring the outputs of wolfram alpha

surreal sierra
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This is sorted bro

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I just wanted to know the difference between assumed zero and real zero

sharp bluff
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wdym assumed 0?

surreal sierra
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Before u said i didn't give enough info to wolfram that's why it assumed as 0..

sharp bluff
surreal sierra
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That's my confusion now

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Like i want to get familiar with it

sharp bluff
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want to get familar with wolfram alpha?

surreal sierra
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Yes

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Ok dis is what i meant

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Difference between these two

wolf-3.png
sharp bluff
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those are two entirely different expressions

surreal sierra
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Ignore the questions the output both r giving 0 ryt?

sharp bluff
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no, the result is at the bottom

surreal sierra
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Evertime?

sharp bluff
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yes every time

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ok, when you say "solve (insert expression here)", wolfram alpha interprets what you mean because it doesnt know what to solve for, thus it assumes your looking for the roots of the expression. you need to specify "solve for (insert varible here) in (insert expression here)"

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for example,

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,w solve for x in log(x)+log(1/x)

daring ravineBOT
sharp bluff
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there, it solve for x and we got x>0

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so when you have log_b(a)+log_b(1/a) and say solve, it doesnt know what variable you want to solve for

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,w solve for a in log_b(a)+log_b(1/a)

daring ravineBOT
sharp bluff
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notice how we got a>0, same as we got x>0

surreal sierra
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Den wt will be the solution of expression which i mistakenly asked in the forum?

sharp bluff
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the expression being log(x)*1/log(x)?

surreal sierra
sharp bluff
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only imaginary solutions

surreal sierra
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Bro ig i fucked up the whole thing starting from wt i joted down in my formula book

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, w solve for a in log_b(a)*log_b(1/a)

daring ravineBOT
surreal sierra
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, w solve for a in log_b(a)*1/log_b(a)

daring ravineBOT
surreal sierra
sharp bluff
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my best advice would be to ditch the calculators and figure this out algebraically. i gtg now so good luck 👍

sharp bluff
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,w simplify log_b(a)*1/log_b(a)

daring ravineBOT
sharp bluff
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now it gives 1

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but i reiterate, ditch the calculators. they are not good for mathematical development and often lead to confusion like what happened here

surreal sierra
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Like wt does basically solve and simplify does

sharp bluff
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solve means you are solving for a variable, simplify means you are just making an expression simpler

wide spear
sharp bluff
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yes it is

wide spear
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what is the issue here

surreal sierra
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.solved