How do you solve $4^{x} = 6^{x + 2}$ $4^x=6^(x+2)$ ?

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Answer 1 · 2016-03-24T07:33:40

In $4^{x} = 6^{x + 2}$ $4^x=6^(x+2)$ , taking log of both the sides, we get

$x \times log 4 = (x + 2) \times log 6$ $x xx log4=(x+2)xxlog6$ or

$x (log 4 - log 6) = 2 log 6$ $x(log4-log6)=2log6$ or

$x = \frac{2 log 6}{log 4 - log 6}$ $x=(2log6)/(log4-log6)$

Now we can simplify using logarithmic tables.

As $log 4 = 0.6021$ $log4=0.6021$ and $log 6 = 0.7782$ $log6=0.7782$

$x = \frac{2 \times 0.7782}{0.6021 - 0.7782}$ $x=(2xx0.7782)/(0.6021-0.7782)$ or

$x = \frac{1.5564}{- 0.1761} = - 8.838$ $x=1.5564/-0.1761=-8.838$

How do you solve 4^x=6^(x+2)4x=6x+24^x=6^(x+2)?