Question

what is the solution of this exponential function?$( \sqrt{2}) ^{x+1} =8 ^{-3x}$

Answer 1

Answer:

${ \sqrt{2} }^{x + 1} = { (\sqrt{2} }^{2} )^{3} )^{ - 3x} \\ { \sqrt{2} }^{x + 1} = \sqrt{2}^{ - 18x} \\ x + 1 = - 18x \\ x = \frac{ - 1}{19}$

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