# Thread: Help with an exponential/logarithmic equation

1. ## Help with an exponential/logarithmic equation

I am trying to solve for x in the equation:

2(5)^x = 3^(x+1)

I know the answer should be x=0.79375 because I've graphed them as functions and found the intersect, as well as run the equation through wolfram.

However, I try to find this by solving for myself and I always seem to come up with x=0.51815.

I have:
2(5)^x = 3^(x+1)

xLog(25) = (x+1)Log3

xLog25 = xLog3 + Log3

xLog25 - xLog3 = Log3

x(Log25 - Log3) = Log3

x(Log25 - Log3)/(Log25 - Log3) = Log3/(Log25 - Log3)

x = Log3/(Log25 - Log3) which is 0.51815

I must be making a mistake somewhere but I just don't see it. Any help is appreciated, thanks!

2. ## Re: Help with an exponential/logarithmic equation

Originally Posted by TypicalNinja
I have:
2(5)^x = 3^(x+1)

xLog(25) = (x+1)Log3
Where did the 25 come from? Was it $\displaystyle \ln 2\times 5^x = x\ln 2\times 5 = x\ln 25$ ?

3. ## Re: Help with an exponential/logarithmic equation

Originally Posted by pickslides
Where did the 25 come from? Was it $\displaystyle \ln 2\times 5^x = x\ln 2\times 5 = x\ln 25$ ?

No, the equation given to me was simply 2 x 5^x = 3^(X+1), I put both side into logarithmic form to try and solve it (context is that this is what I'm suppose to do for every question in this assignment).

I skipped out a step when describing what I did, my first step was to log both sides, so the equation became 2log5^x = log3^(x+1). Then I used the power law to take the exponent of the logs and change it to the coefficient. So the equation became 2xlog5 = (X+1)log3. And I also did the reverse on the left side, took the coefficient 2 and brought it back into the log, giving me xlog5^2, which is xlog25. Am I making a mistake somewhere in that process?

4. ## Re: Help with an exponential/logarithmic equation

Taking the logs of both sides is a good start, so you get:
$\displaystyle \log(2\cdot 5^x)=\log(3^{x+1})$
$\displaystyle \Leftrightarrow \log(2)+x\log(5)=(x+1)\log(3)$
Bring the variable(s) $\displaystyle x$ and the constants to one side.

5. ## Re: Help with an exponential/logarithmic equation

deepashree k s

yes, u did a mistake in first step.

u take log for 2(5)^x that becomes log2+xlog5.by applying this u slove the problem and you can get ur result 0.7937.

6. ## Re: Help with an exponential/logarithmic equation

Originally Posted by Siron
Taking the logs of both sides is a good start, so you get:
$\displaystyle \log(2\cdot 5^x)=\log(3^{x+1})$
$\displaystyle \Leftrightarrow \log(2)+x\log(5)=(x+1)\log(3)$
Bring the variable(s) $\displaystyle x$ and the constants to one side.
Oh I see, my mistake was not applying the log to the entire left side of the equation (I only applied the log to the 5^x, and left the 2 alone during this step). Basic algebra error, I had a feeling I was messing up something basic like that. Thanks very much!

EDIT: Thanks deepashree as well!