# finding the vaule of a raised power

• May 20th 2012, 10:26 AM
davellew69
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
• May 20th 2012, 10:45 AM
earboth
Re: finding the vaule of a raised power
Quote:

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:

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

• May 20th 2012, 10:48 AM
davellew69
Re: finding the vaule of a raised power
can this be done by using algebra?
• May 20th 2012, 01:14 PM
Kanderson
Re: finding the vaule of a raised power
That is algebra.
• May 20th 2012, 02:55 PM
bjhopper
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
• May 20th 2012, 06:35 PM
Wilmer
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!