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\[y=3^{x}\]
\[y=_{5}x\]

yep, I did that.

what you can do is switch your x's and y's then just solve for y.

I have answers but I want to know what I did wrong, because it seems right to me.

i think you need to keep the logs or put e^y

you can only take away the logs like you did if y had one on that side

i think it can be rewrote as y=ln(x-3) but i'm a little rusty with logarithms

but there is log10 on both sides, so I can cancel it out? and ok... I'm still confused sorry.

when you moved the y over to the other sice you had y=log(x)/log(3) y doesn't have a log

if it was log(y)=log(x)/log(3) you could cancel them out

really? hmmm, confusing but I know what you mean.

yeah i get confused too when you start using logs

the answer is \[\log _{3}x for x>0\]
can you help me out? I don't understand how they got it.

log 2 x or log 3 x?

\[\log_{a}y=x \]
is \[a^{x}=y\]

3x

from log10(x)=ylog10(3)
can you show me the rest of the steps?

so we can rewriteyour first one as as\[\log_{3} x=y\]

after you switch the x and y I believe?

sorry i'm trying to remember this as we go

it's ok.

no that is just rewriting it then we swith and solve for x

you wrote it wrong then.

all i did was change the exponent to a log

\[y=3^{x}\]
\[\log _{3}y=x\]

I think you got confused with the x and y, you just got them in the wrong position.

then we can switch the x & y from there.

it's hard to do math on these boards lol

yeh I know haha, especially trying to write the equation takes years.

using the equation tab*

i think i have it let me check with my calculator

kk.

by the rules of logarithms we have loga^b=c is the same as a^c=b

ok, and it turns out to be\[y=\log _{3}x\] right?

yes.

so let's try the second one

y=5^x is log5^y=x so swap is log5^x=y

yep but the answer only says 5^x for some reason.

confusing, I'll just have to ask my teacher. I'm just doing it advanced anyway.

thanks for the help.