hey,

im stuck on this question i don't know how to takle it.

3(2^3X) = 5(4^2X-3)

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- Dec 17th 2011, 07:56 PMwaleedrabbaniSolve for x (Exponential)
hey,

im stuck on this question i don't know how to takle it.

3(2^3X) = 5(4^2X-3) - Dec 17th 2011, 08:27 PMTKHunnyRe: Solve for x (Exponential)
$\displaystyle 4^{2x-3} = (2^{2})^{2x-3} = 2^{4x-6}$

It's a secret. Not the same Base? Make it the same base and your life WILL be easier. Whoops, I let the secret out. - Dec 20th 2011, 04:03 AMiragequitRe: Solve for x (Exponential)
3(2^3X) = 5(4^2X-3)

if by 5(4^2x-3) you mean 5(4^(2x-3))

take common log of both sides, use product law of logarithms, use power law for logarithms (to bring the exponents down), factor x, and you should get x = (log5-log3-3log4)/(3log2-2log4) as an exact answer

if by 5(4^2x-3) you mean what you say, then i got x= (log15log3 - log5) / (2log4-3log2). method is the same as above