How do you solve $2^x=sqrt(3^(x-2))$ ?

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Answer 1 · 2016-11-08T14:16:04

Square both sides.

$(2^x)^2 = (sqrt(3^(x- 2)))^2$

$2^(2x) = 3^(x- 2)$

Take the natural logarithm of both sides.

$ln(2^(2x)) = ln(3^(x- 2))$

Use the power rule of logarithms, $ln(a^n) = nlna$

$(2x)ln2 = (x - 2)ln3$

$2xln2 = xln3 - 2ln3$

$2xln2 - xln3 = -2ln3$

$x(2ln2 - ln3) = -2ln3$

You can simplify the expression in parentheses using the rules $nlna = lna^n$ and $lna - lnb = ln(a/b)$ .

$x(ln4 - ln3) = -2ln3$

$x(ln4/3) = ln(1/9)$

$x = ln(1/9)/(ln(4/3))$

$x~= -7.64$

Hopefully this helps!

How do you solve 2^x=sqrt(3^(x-2))2^x=sqrt(3^(x-2))?