# Thread: finding the vaule of a raised power

1. ## finding the vaule of a raised power

I have the following

12.0*1.08^x =25.0

I need to find the value of x but Im lost on how its done.

Dave

2. ## Re: finding the vaule of a raised power

Originally Posted by davellew69
I have the following

12.0*1.08^x =25.0

I need to find the value of x but Im lost on how its done.

Dave
1. Divide through by 12.0.

2. Actually you have an exponential term at the LHS so transform the RHS into an exponential term with the same base:

$\displaystyle{1.08^x = 1.08^{\log_{1.08}\left(\frac{25}{12} \right)}}$ ............ So $x = \log_{1.08}\left(\frac{25}{12} \right)$

3. To get an approximate value for x use the base change formula:

$\log_a(b)=\frac{\ln(b)}{\ln(a)}$

3. ## Re: finding the vaule of a raised power

can this be done by using algebra?

4. ## Re: finding the vaule of a raised power

That is algebra.

5. ## Re: finding the vaule of a raised power

I would simply use log base 10. No need to change.
log 12 + x(log 1.08) = log 25
x= 9.5

6. ## Re: finding the vaule of a raised power

1.08^x = 25/12

x = log(25/12) / log(1.08)

If a^x = b then x = log(b) / log(a) : tattoo that on your wrist!