How do you solve $3^{2 x} = 4^{x + 1}$ $3^(2x) = 4^(x+1)$ ?

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Answer 1 · 2015-09-18T15:54:16

$x = \frac{log (2)}{log (3) - log (2)}$ $x = (log(2))/(log(3) - log(2))$

Take the log of both sides
$log (3^{2 x}) = log (4^{x + 1})$ $log(3^(2x)) = log(4^(x+1))$

Remember the property that $log (m^{n}) = n log (m)$ $log(m^n) = nlog(m)$

$2 x log (3) = (x + 1) log (4)$ $2xlog(3) = (x+1)log(4)$

Expand

$2 x log (3) = x log (4) + log (4)$ $2xlog(3) = xlog(4) + log(4)$

Put everything with an x to the same side

$2 x log (3) - x log (4) = log (4)$ $2xlog(3) - xlog(4) = log(4)$

Put x on evidence, and put that 2 back in the log

$x (2 log (3) - log (4)) = log (4)$ $x(2log(3) - log(4)) = log(4)$

Isolate x

$x = \frac{log (4)}{2 log (3) - log (4)}$ $x = log(4)/(2log(3) - log(4))$

Remember that $4 = 2^{2}$ $4 = 2^2$ , so #log(4) = 2log(2)

$x = \frac{2 log (2)}{2 log (3) - 2 log (2)}$ $x = (2log(2))/(2log(3) - 2log(2))$

Put 2 in evidence and cancel it

$x = \frac{log (2)}{log (3) - log (2)}$ $x = (log(2))/(log(3) - log(2))$

How do you solve 32x=4x+1 3^(2x) = 4^(x+1)?