How do you solve $6^{x - 3} = 2^{x}$ $6^(x-3) = 2^x$ ?

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Answer 1 · 2016-05-04T16:13:37

$x = 3 \frac{log 6}{log 6 - log 2} = 4.893$ $x=3 log 6/(log 6-log 2)=4.893$ , nearly.

Equating logarithms,

$(x - 3) log 6 = x log 2$ $(x-3)log 6=x log 2$ . Solving for x.

$x = 3 \frac{log 6}{log 6 - log 2}$ $x=3 log 6/(log 6-log 2)$

How do you solve 6x−3=2x 6^(x-3) = 2^x?