How do you solve $12^{x - 4} = 3^{x - 2}$ $12^(x-4)=3^(x-2)$ ?

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Answer 1 · 2017-01-07T14:01:34

$x = 5.585$ $x=5.585$

As $12^{x - 4} = 3^{x - 2}$ $12^(x-4)=3^(x-2)$ , taking log on both sides, we get

$(x - 4) log 12 = (x - 2) log 3$ $(x-4)log12=(x-2)log3$

or $x log 12 - 4 log 12 = x log 3 - 2 log 3$ $xlog12-4log12=xlog3-2log3$

or $x log 12 - x log 3 = 4 log 12 - 2 log 3$ $xlog12-xlog3=4log12-2log3$

or $x (log 12 - log 3) = log (\frac{12^{4}}{3^{2}})$ $x(log12-log3)=log((12^4)/(3^2))$

or $x log 4 = log (4^{2} \cdot 12^{2}) = 2 log 48$ $xlog4=log(4^2*12^2)=2log48$

or $x = \frac{2 log 48}{log 4} = \frac{2 \times 1.6812}{0.6021} = 5.585$ $x=(2log48)/(log4)=(2xx1.6812)/0.6021=5.585$

How do you solve 12^(x-4)=3^(x-2)12x−4=3x−212^(x-4)=3^(x-2)?