Question

5^x=3^(x-1)
solve for x.

Answer 1

I would multiply both sides by 3 to get rid of the (x-1):
\[3*5^x=3*3^{x-1}=3^x\]
Now use the property of logarithms that \[\ln {a^b}= b \ln {a}\] to take the log of both sides
\[\ln{(3*5^x)}=\ln{3^x}=x\ln{3}\]
Now use the property of logarithms that \[\ln{(a*b)}=\ln{a}*\ln{b}\] to simplify the left side
\[\ln{3}+\ln{5^x} = x \ln{3}\]\[\ln{3}+x\ln{5} = x \ln{3}\]
\[\ln{3}=x\ln{3}-x\ln{5} = x(\ln{3}-\ln{5})\]\[x=\frac{\ln{3}}{\ln{3}-\ln{5}}\]

Answer 2

I didn't get the first step.

Answer 3

Wouldn't it become 15^x=9^(x-1)

Answer 4

No. \[5^x = 5*5*5*5...\] (x 5's)
\[3*5^x= 3*5*5*5*5...\] (x 5's)

Answer 5

and \[3*3^{x-1} = 3*(3*3*3*3*...)\] (x-1 3's in the parentheses)

Answer 6

Remember \[x^a*x^b=x^{a+b}\]and\[3^1=3\]

Answer 7

Thankyou.

Answer 8

\[5^x=3^{(x-1)}\]\[5^x = 3^x \dot\ 3^{-1}\]\[5^x = 3^x \dot\ \frac{1}{3}\]\[\ \frac{5^x}{3^x} = \frac{1}{3}\]\[\left(\frac{5}{3}\right)^x = \frac{1}{3}\]

That makes things easier. Don't deal with logs until you absolutely have to.

Answer 9

Just did another problem with logs, got logs on the brain :-)

Answer 10

I personally like to hold off logs as long as I possibly can.

Answer 11

Seeing ln's all over hurts my eyes

Answer 12

Well, if nothing else, your route might be easier to typeset here...

Answer 13

I just realized I could have expressed that even simpler than that.

Answer 14

I could have expressed it as

\[3 = \left(\frac{3}{5}\right)^x\]

Answer 15

Yeah, I like that. Quick trip from there to the answer.

Answer 16

Yup, that's exactly what I aim for.

Answer 17

Thanks mate.

Answer 18

I'm here to make your life easier

Answer 19

As you can see, I hate logs