JayDS
Find the equation of the inverse of the following.
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JayDS
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\[y=3^{x}\]
\[y=_{5}x\]
Jusaquikie
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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.
JayDS
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yep, I did that.
Jusaquikie
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what you can do is switch your x's and y's then just solve for y.
JayDS
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\[y=3^{x}\]
\[x=3^{y}\]
\[\log _{10}x = \log _{10}3^{y}\]
\[\log _{10} x = ylog _{10}3\]
\[y=\frac{\log _{10}x}{\log _{10}3}\]
y=x/3?
JayDS
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I have answers but I want to know what I did wrong, because it seems right to me.
Jusaquikie
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i think you need to keep the logs or put e^y
Jusaquikie
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you can only take away the logs like you did if y had one on that side
Jusaquikie
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i think it can be rewrote as y=ln(x-3) but i'm a little rusty with logarithms
JayDS
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but there is log10 on both sides, so I can cancel it out? and ok... I'm still confused sorry.
Jusaquikie
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when you moved the y over to the other sice you had y=log(x)/log(3) y doesn't have a log
Jusaquikie
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if it was log(y)=log(x)/log(3) you could cancel them out
JayDS
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really? hmmm, confusing but I know what you mean.
Jusaquikie
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yeah i get confused too when you start using logs
Jusaquikie
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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.
JayDS
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the answer is \[\log _{3}x for x>0\]
can you help me out? I don't understand how they got it.
Jusaquikie
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log 2 x or log 3 x?
Jusaquikie
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\[\log_{a}y=x \]
is \[a^{x}=y\]
JayDS
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3x
JayDS
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from log10(x)=ylog10(3)
can you show me the rest of the steps?
Jusaquikie
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so we can rewriteyour first one as as\[\log_{3} x=y\]
JayDS
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after you switch the x and y I believe?
Jusaquikie
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sorry i'm trying to remember this as we go
JayDS
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it's ok.
Jusaquikie
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no that is just rewriting it then we swith and solve for x
JayDS
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you wrote it wrong then.
Jusaquikie
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all i did was change the exponent to a log
JayDS
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\[y=3^{x}\]
\[\log _{3}y=x\]
JayDS
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I think you got confused with the x and y, you just got them in the wrong position.
JayDS
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then we can switch the x & y from there.
Jusaquikie
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it's hard to do math on these boards lol
JayDS
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yeh I know haha, especially trying to write the equation takes years.
JayDS
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using the equation tab*
Jusaquikie
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i think i have it let me check with my calculator
JayDS
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kk.
Jusaquikie
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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
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by the rules of logarithms we have loga^b=c is the same as a^c=b
JayDS
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ok, and it turns out to be\[y=\log _{3}x\] right?
Jusaquikie
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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?
JayDS
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yes.
Jusaquikie
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so let's try the second one
Jusaquikie
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y=5^x is log5^y=x so swap is log5^x=y
JayDS
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yep but the answer only says 5^x for some reason.
JayDS
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confusing, I'll just have to ask my teacher. I'm just doing it advanced anyway.
JayDS
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thanks for the help.