Find the equation of the inverse of the following.

- anonymous

Find the equation of the inverse of the following.

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- anonymous

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

- Jusaquikie

the inverse is just if you put in an x and get a y the inverse is like putting in the y and getting the same x you put in the first part.

- anonymous

yep, I did that.

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## More answers

- Jusaquikie

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

- anonymous

\[y=3^{x}\]
\[x=3^{y}\]
\[\log _{10}x = \log _{10}3^{y}\]
\[\log _{10} x = ylog _{10}3\]
\[y=\frac{\log _{10}x}{\log _{10}3}\]
y=x/3?

- anonymous

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

- Jusaquikie

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

- Jusaquikie

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

- Jusaquikie

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

- anonymous

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

- Jusaquikie

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

- Jusaquikie

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

- anonymous

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

- Jusaquikie

yeah i get confused too when you start using logs

- Jusaquikie

if you have the answer try to work it backwards to get the first question and try to tie the two to gether with what your sure of.

- anonymous

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

- Jusaquikie

log 2 x or log 3 x?

- Jusaquikie

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

- anonymous

3x

- anonymous

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

- Jusaquikie

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

- anonymous

after you switch the x and y I believe?

- Jusaquikie

sorry i'm trying to remember this as we go

- anonymous

it's ok.

- Jusaquikie

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

- anonymous

you wrote it wrong then.

- Jusaquikie

all i did was change the exponent to a log

- anonymous

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

- anonymous

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

- anonymous

then we can switch the x & y from there.

- Jusaquikie

it's hard to do math on these boards lol

- anonymous

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

- anonymous

using the equation tab*

- Jusaquikie

i think i have it let me check with my calculator

- anonymous

kk.

- Jusaquikie

ok lol i can't figure out how to do it on my calculator but i think it's right, so all we are doing is changing the question into a log of the same equasion then when we swap x and y we already have the answer

- Jusaquikie

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

- anonymous

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

- Jusaquikie

so taking your problem y=3^x and putting it in log form we get log3^y=x then we swap y and x to get log3^x=y is that right?

- anonymous

yes.

- Jusaquikie

so let's try the second one

- Jusaquikie

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

- anonymous

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

- anonymous

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

- anonymous

thanks for the help.

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