anonymous
  • anonymous
Find the equation of the inverse of the following.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
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anonymous
  • anonymous
\[y=3^{x}\] \[y=_{5}x\]
Jusaquikie
  • 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
  • anonymous
yep, I did that.

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More answers

Jusaquikie
  • Jusaquikie
what you can do is switch your x's and y's then just solve for y.
anonymous
  • 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
  • anonymous
I have answers but I want to know what I did wrong, because it seems right to me.
Jusaquikie
  • Jusaquikie
i think you need to keep the logs or put e^y
Jusaquikie
  • Jusaquikie
you can only take away the logs like you did if y had one on that side
Jusaquikie
  • Jusaquikie
i think it can be rewrote as y=ln(x-3) but i'm a little rusty with logarithms
anonymous
  • anonymous
but there is log10 on both sides, so I can cancel it out? and ok... I'm still confused sorry.
Jusaquikie
  • 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
  • Jusaquikie
if it was log(y)=log(x)/log(3) you could cancel them out
anonymous
  • anonymous
really? hmmm, confusing but I know what you mean.
Jusaquikie
  • Jusaquikie
yeah i get confused too when you start using logs
Jusaquikie
  • 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
  • anonymous
the answer is \[\log _{3}x for x>0\] can you help me out? I don't understand how they got it.
Jusaquikie
  • Jusaquikie
log 2 x or log 3 x?
Jusaquikie
  • Jusaquikie
\[\log_{a}y=x \] is \[a^{x}=y\]
anonymous
  • anonymous
3x
anonymous
  • anonymous
from log10(x)=ylog10(3) can you show me the rest of the steps?
Jusaquikie
  • Jusaquikie
so we can rewriteyour first one as as\[\log_{3} x=y\]
anonymous
  • anonymous
after you switch the x and y I believe?
Jusaquikie
  • Jusaquikie
sorry i'm trying to remember this as we go
anonymous
  • anonymous
it's ok.
Jusaquikie
  • Jusaquikie
no that is just rewriting it then we swith and solve for x
anonymous
  • anonymous
you wrote it wrong then.
Jusaquikie
  • Jusaquikie
all i did was change the exponent to a log
anonymous
  • anonymous
\[y=3^{x}\] \[\log _{3}y=x\]
anonymous
  • anonymous
I think you got confused with the x and y, you just got them in the wrong position.
anonymous
  • anonymous
then we can switch the x & y from there.
Jusaquikie
  • Jusaquikie
it's hard to do math on these boards lol
anonymous
  • anonymous
yeh I know haha, especially trying to write the equation takes years.
anonymous
  • anonymous
using the equation tab*
Jusaquikie
  • Jusaquikie
i think i have it let me check with my calculator
anonymous
  • anonymous
kk.
Jusaquikie
  • 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
  • Jusaquikie
by the rules of logarithms we have loga^b=c is the same as a^c=b
anonymous
  • anonymous
ok, and it turns out to be\[y=\log _{3}x\] right?
Jusaquikie
  • 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
  • anonymous
yes.
Jusaquikie
  • Jusaquikie
so let's try the second one
Jusaquikie
  • Jusaquikie
y=5^x is log5^y=x so swap is log5^x=y
anonymous
  • anonymous
yep but the answer only says 5^x for some reason.
anonymous
  • anonymous
confusing, I'll just have to ask my teacher. I'm just doing it advanced anyway.
anonymous
  • anonymous
thanks for the help.

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