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Find the equation of the inverse of the following.

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\[y=3^{x}\] \[y=_{5}x\]
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.
yep, I did that.

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Other answers:

what you can do is switch your x's and y's then just solve for y.
\[y=3^{x}\] \[x=3^{y}\] \[\log _{10}x = \log _{10}3^{y}\] \[\log _{10} x = ylog _{10}3\] \[y=\frac{\log _{10}x}{\log _{10}3}\] y=x/3?
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
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.
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 \[a^{x}=y\]
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
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
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?
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?
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.

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