Find the equation of the inverse of the following.

Question

anonymous · Answer

$$y=3^{x}$$

$$y=_{5}x$$

jusaquikie · Answer

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 · Answer

yep, I did that.

jusaquikie · Answer

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

anonymous · Answer

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

anonymous · Answer

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

jusaquikie · Answer

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

jusaquikie · Answer

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

jusaquikie · Answer

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

anonymous · Answer

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

jusaquikie · Answer

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

jusaquikie · Answer

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

anonymous · Answer

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

jusaquikie · Answer

yeah i get confused too when you start using logs

jusaquikie · Answer

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 · Answer

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

jusaquikie · Answer

log 2 x or log 3 x?

jusaquikie · Answer

$$\log_{a}y=x $$
is $$a^{x}=y$$

anonymous · Answer

3x

anonymous · Answer

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

jusaquikie · Answer

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

anonymous · Answer

after you switch the x and y I believe?

jusaquikie · Answer

sorry i'm trying to remember this as we go

anonymous · Answer

it's ok.

jusaquikie · Answer

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

anonymous · Answer

you wrote it wrong then.

jusaquikie · Answer

all i did was change the exponent to a log

anonymous · Answer

$$y=3^{x}$$
$$\log _{3}y=x$$

anonymous · Answer

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

anonymous · Answer

then we can switch the x & y from there.

jusaquikie · Answer

it's hard to do math on these boards lol

anonymous · Answer

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

anonymous · Answer

using the equation tab*

jusaquikie · Answer

i think i have it let me check with my calculator

anonymous · Answer

kk.

jusaquikie · Answer

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 · Answer

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

anonymous · Answer

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

jusaquikie · Answer

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 · Answer

yes.

jusaquikie · Answer

so let's try the second one

jusaquikie · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

thanks for the help.