# Solving exponential to different bases and a constant term

• Mar 19th 2009, 07:46 PM
skylabel
Solving exponential to different bases and a constant term
Trying to solve for x
2^(0.5x)= [3^(-0.5x)] + 5/3

I know, the usual, take logs of both sides and bring down the exponent, but I'm having a problem with the RHS of the equation, since I can't take logs seperately, i.e. log(a+b) is NOT equal to log(a)+log(b)

I actually know (graphically) that the answer is x=2, and I know that's correct from substituting. Any ideas on how to approach it (besides trial and error)
• Mar 20th 2009, 03:13 AM
mr fantastic
Quote:

Originally Posted by skylabel
Trying to solve for x
2^(0.5x)= [3^(-0.5x)] + 5/3

I know, the usual, take logs of both sides and bring down the exponent, but I'm having a problem with the RHS of the equation, since I can't take logs seperately, i.e. log(a+b) is NOT equal to log(a)+log(b)

I actually know (graphically) that the answer is x=2, and I know that's correct from substituting. Any ideas on how to approach it (besides trial and error)

If you know that there's a solution and that you have to find it by hand, it makes sense to guess-and-check even values of x. And x = 2 would naturally be the first value to check .....

Off-hand, I don't see any simple algebraic approach to the question.
• Mar 20th 2009, 04:54 AM
Hush_Hush
Hello!

If you will work with the formula a little you will be able to write it like this:

$\displaystyle (\sqrt3)^x * [(\sqrt2)^x -5/3 ]=1$ (1)

as you can observe $\displaystyle (\sqrt3)^x > 0$=>$\displaystyle (\sqrt2)^x -5/3 >0$ =>

$\displaystyle (\sqrt2)^x > 5/3 => x>1,5$ (2)

The second step is to observe that with x>1 => $\displaystyle (\sqrt3)^x >1$ => $\displaystyle (\sqrt2)^x -5/3 < 1$ in order for equation (1) to be verified.

From the condition $\displaystyle (\sqrt2)^x -5/3 < 1$=> $\displaystyle (\sqrt2)^x<8/3$ => x<2.5 (3)

From the condition (2) and (3) we have 1 < x< 3 . The only solution for x in this interval in order for $\displaystyle (\sqrt2)^x$ and $\displaystyle (\sqrt3)^x$ to be rational numbers is x=2;

To demonstrate that this is the only solution it,s easy to study the monotony of the function in interval ( 1,5 ; 2,5)

Have a nice day!