logarithm

[kindred kernel]

need answer

[pliant wave]

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[tame valve]

simplify lhs using base change property of logs

the property is $log_{b}{a} = \frac{\log{a}}{\log{b}}$ if youre wondering

[tawny pulsar]

ultra racism | #GTFOanemia

[kindred kernel]

i tried using this property

but there would be 3bases then

x, 2x, y

[tame valve]

just convert everything into base 10

[kindred kernel]

can you solve it for me?

[tame valve]

$\log_{3}x = \frac{\log{x}}{\log{3}}$

[tawny pulsar]

ultra racism | #GTFOanemia

[kindred kernel]

i am getting y = 5

wrong asnwer

[tame valve]

$\log_{x}{2x} = \frac{\log{2x}}{\log{x}}$

[tawny pulsar]

ultra racism | #GTFOanemia

[kindred kernel]

$\log_{2x}{y} = \frac{\log{y}}{\log{2x}}$

[tawny pulsar]

ghoul

[tame valve]

yep

now multiply them out

[kindred kernel]

now
1/log3 * log y = 2

[tame valve]

ye

[kindred kernel]

log ( y - 3 ) = 2

[tame valve]

so y=9

noooooo

nooooooo

[kindred kernel]

what is it?

[tame valve]

$\frac{\log{y}}{\log{3}} = 2$

[tawny pulsar]

ultra racism | #GTFOanemia

[tame valve]

$\log{y} = 2\log{3}$

[tawny pulsar]

ultra racism | #GTFOanemia

[kindred kernel]

can we not take log common?

and multiply base 10

[tame valve]

$\log{y} = \log{3^2} = \log{9}$

[tawny pulsar]

ultra racism | #GTFOanemia

[tame valve]

so y=9

[kindred kernel]

i get it, but can we not take log common

from log ( y - 3) = 2

y-3 = 10^2

[tame valve]

so youre saying like, $\log{y-3} = 2$?

[tawny pulsar]

ultra racism | #GTFOanemia

[kindred kernel]

$log{(y-3)} = 2$

[tawny pulsar]

ghoul

[kindred kernel]

$(y-3) = \exp 10^2$

[tame valve]

you are using the property wrong bro

[tawny pulsar]

ghoul

[tame valve]

@kindred kernel that is not a property

$\log{a}-\log{b} = \log{\frac{a}{b}}$

[tawny pulsar]

ultra racism | #GTFOanemia

[tame valve]

$\frac{\log{a}}{\log{b}} \neq \log{(a-b)}$

@kindred kernel

[tawny pulsar]

ultra racism | #GTFOanemia

[kindred kernel]

ah got it

[tame valve]

is it clarified now

[kindred kernel]

is it correct?
$\log{a} + \log{b} = \log{{a}{b}}$

[tawny pulsar]

ghoul

[tame valve]

ye

[kindred kernel]

and this?

log a * log b = log a+b

[tame valve]

no

[kindred kernel]

edited*

[tame valve]

no still false

[kindred kernel]

ah okk

[tame valve]

alr is it clear now

[kindred kernel]

yes

thanks

[tame valve]

if u feel i helped then nominate me in #helper-nominations and close the thread

[kindred kernel]

+close