How do you solve $.5 (6^{x}) = 2^{x}$ $.5(6^x) = 2^x$ ?

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Answer 1 · 2016-01-13T19:35:53

$x = 0.631$ $x=0.631$ (3sf)

Take logs of both sides.

$log (0.5 \cdot (6^{x})) = log (2^{x})$ $log(0.5*(6^x))=log(2^x)$

$log 0.5 + x log 6 = x log 2$ $log0.5 + xlog6 = xlog2$

$log 0.5 = x log 2 - x log 6$ $log0.5=xlog2-xlog6$

$log 0.5 = x (log 2 - log 6)$ $log0.5=x(log2-log6)$

$x = \frac{log 0.5}{log 2 - log 6}$ $x=(log0.5)/(log2-log6)$

$x = \frac{log 2}{log 6 - log 2}$ $x=log2/(log6-log2)$

$x = \frac{log 2}{log 3}$ $x=log2/log3$

$x = 0.631$ $x=0.631$ (3sf)

How do you solve .5(6x)=2x.5(6^x) = 2^x?