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!