can someone tell me where i went wrong ?

[long dagger]

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[cobalt narwhal]

can u write down the

[burnt yarrow]

hey

[cobalt narwhal]

problem

in text

pls

[burnt yarrow]

wdym in text

[cobalt narwhal]

like

in chat

[burnt yarrow]

alr

2 log x = 1 + log (x-1)

its in log base 4

[simple cloak]

bro it's correct-

[burnt yarrow]

huh

correct answer is 2

[cobalt narwhal]

whats the

base

[burnt yarrow]

4

[cobalt narwhal]

oh

4

[burnt yarrow]

i got 20/3 but its 2

[cobalt narwhal]

x = 2

is coming

for me

is that correct

[burnt yarrow]

yes but whats wrong w my method

it is correct

2 is the ans but why am i getting 20/3

[simple cloak]

huh wait a bit

[burnt yarrow]

guys my doubt lies in what went wrong in what i did

[cobalt narwhal]

here

when u add like that u didn't

put base as

x

BUT AS 4

bro

🗿

log_x (4x-4)

is how u put it

like

when u divide like that

my method is better btw..

[simple cloak]

OH YEAH-

[burnt yarrow]

oh

[cobalt narwhal]

he put 4 kek

[burnt yarrow]

yeah i see it

thanks

[simple cloak]

subtracting logarithms results in division inside the log-

[cobalt narwhal]

yeh

[simple cloak]

dividing them turns them into that-

bruhhhh

I need to brush up on logarithms as well-

wait-

[cobalt narwhal]

just take them one side and do it that way why even divide x will go into power LOL thats more fucking annoying

to solve

💀

[burnt yarrow]

ty guys

[cobalt narwhal]

this will put x on top of a constant

2

🗿

lmao

thats not

solving

anything

[simple cloak]

$2\log_4(x) = 1+\log_4(x-1)$

$4^2 = 4(x-1)$

[fading sleet]

Ravi #GTFOanemia#NoLifer

[simple cloak]

would this work?

just take everything as an exponent of 4

no...

we get =5

[burnt yarrow]

is there a rule for log a/ log b

[simple cloak]

yeah i think there is

[burnt yarrow]

oh

do you know what it is?

[simple cloak]

hmmm

we can try deriving it

try

$\ln(e^{\frac{\ln(a)}{\ln(b)}})$

[fading sleet]

Ravi #GTFOanemia#NoLifer

[cobalt narwhal]

bro took it to

another level

[simple cloak]

yeah idk how to derive this law

um

[cobalt narwhal]

which?

[simple cloak]

the log division one

[cobalt narwhal]

oh i know that

one

[simple cloak]

wait...

[cobalt narwhal]

should i do?

[simple cloak]

$\log_a(b)$ = \log_a(c^{\log_c(b)}) = \log_a(c)\log_c(b)$

$\log_a(b) = \log_a(c)\log_c(b)$

$\frac{\log_a(b)}{\log_a(c)} = \log_c(b)$

[fading sleet]

Ravi #GTFOanemia#NoLifer
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[simple cloak]

well that worked

[cobalt narwhal]

take the things into exponents then take a new log with different base

and put them in

ratio

yeh

😴

[simple cloak]

I've done this trick with natural log before tbh

[cobalt narwhal]

i learnt this in school

[simple cloak]

I'm in school rn lol

we haven't learnt log yet

[simple cloak]

(like I'm studying in school)

I'm not currently at school

[cobalt narwhal]

bro

lmao

school taught log in

grade 5

[simple cloak]

wait what