# Exponential Equations.

• Feb 25th 2010, 04:38 AM
aligator0207
Exponential Equations.
Solution to the exponential equation:

3^2x-1=5

Solve the logarithmic equation for X:

ln(2 + X) =1
• Feb 25th 2010, 04:46 AM
Quote:

Originally Posted by aligator0207
Solution to the exponential equation:

3^2x-1=5

Solve the logarithmic equation for X:

ln(2 + X) =1

hi

(1) $\displaystyle 3^{2x}=6$

$\displaystyle 2x \lg 3=\lg 6$

$\displaystyle 2x=\frac{\lg 6}{\lg 3}$

.....

(2) Write in another form : $\displaystyle e^1=2+x$ , e is a constant with value so now you can solve for x
• Feb 25th 2010, 07:39 AM
aligator0207
What happened to the (-1) in problem 1? And at the end, do I put that into my calculator?
• Feb 25th 2010, 08:02 AM
Hello aligator0207
Quote:

Originally Posted by aligator0207
What happened to the (-1) in problem 1? And at the end, do I put that into my calculator?

I suspect that you should have put brackets around the 2x-1 and that the original equation was actually:
$\displaystyle 3^{2x-1}=5$
and, if so, the working should be:
$\displaystyle (2x-1)\log3=\log5$

$\displaystyle \Rightarrow 2x-1 = \frac{\log5}{\log3}$

$\displaystyle \Rightarrow x = \frac12\left(1+\frac{\log5}{\log3}\right)$
$\displaystyle =\frac{\log3+\log5}{2\log3}$

$\displaystyle =\frac{\log15}{\log9}$

$\displaystyle =1.2325$