Exponential Equations

[Deo3560]

Ok, I did manage to figure out the answer to the problem, but I keep trying to duplicate it and never can get the same answer.

The Problem is:
3^(2x)=13^(x-5)

The answer I managed to get was:
34.876=x

I was just plugging things into the calculator to figure it out, and I did come across the answer, but I cant figure out how >.<.
So if someone could help me that would be awesome.

Thanks,
Deo3560

 

[wjm11]

The Problem is:
3^(2x)=13^(x-5)

1) Take the log of each side.

2) Use the property of logs that says you can move the exponent to the coefficient position.

Should be obvious from there. Hope that helps.

 

[masters]

Ok, I did manage to figure out the answer to the problem, but I keep trying to duplicate it and never can get the same answer.

The Problem is:
3^(2x)=13^(x-5)

The answer I managed to get was:
34.876=x

I was just plugging things into the calculator to figure it out, and I did come across the answer, but I cant figure out how >.<.
So if someone could help me that would be awesome.

Thanks,
Deo3560

Hi Deo3560,

Your answer is correct to 3 rounded decimal places.

Here's the setup algebraically.

\(\displaystyle 3^{2x}=13^{x-5}\)

\(\displaystyle 2x \ln 3=(x-5) \ln 13\)

\(\displaystyle 2x \ln 3=x\ln 13 - 5 \ln 13\)

\(\displaystyle 2x \ln 3 - x \ln 13 = -5 \ln 13\)

\(\displaystyle x(2\ln 3 - \ln 13)=-5 \ln 13\)

\(\displaystyle x=\frac{-5 \ln 13}{2\ln 3 - \ln 13}\)