How do you solve #(\sqrt { 2} ) ^ { x + 5} = 4^ { x }#?

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Answer 1 · 2017-09-13T01:58:33

Given: #(sqrt2) ^ (x + 5) = 4^ ( x )#

Use the base 10 logarithm on both sides:

#log_10((sqrt2) ^ (x + 5)) = log_10(4^ ( x ))#

Logarithms have the property #log_b(a^c) = (c)log_b(a)#; use that property on both sides:

#(x + 5)log_10(sqrt2) = (x)log_10(4)#

Divide both sides by #log_10(sqrt2)#:

#x + 5 = (x)log_10(4)/log_10(sqrt2)#

Use a calculator to perform the division:

#x + 5 = 4x#

Solve for x:

#3x = 5#

#x = 5/3#