How do you solve $3^{x + 4} = 7^{2 x - 1}$ $3^(x+4)=7^(2x-1)$ ?

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How do you solve $3^{x + 4} = 7^{2 x - 1}$ $3^(x+4)=7^(2x-1)$ ?

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Answer 1 · 2018-06-28T14:38:44

Take logarithms

Explanation:

$3^{x + 4} = 7^{2 x - 1}$ $3^(x+4)=7^(2x-1)$

Take logs:

$(x + 4) log 3 = (2 x - 1) log 7$ $(x+4)log3=(2x-1)log7$
$x (log 3 - 2 log 7) = - log 7 - 4 log 3$ $x(log3-2log7)=-log7-4log3$

$x = \frac{- log 7 - 4 log 3}{log 3 - 2 log 7} = \frac{log 7 + 4 log 3}{2 log 7 - log 3}$ $x=(-log7-4log3)/(log3-2log7)=(log7+4log3)/(2log7-log3)$
$= \frac{log 7 + log 81}{log 49 - log 3} = \frac{log 567}{log (\frac{49}{3})} = {log}_{\frac{49}{3}} 567$ $=(log7+log81)/(log49-log3)=log567/log(49/3)=log_(49/3)567$

To three significant figures, $x = 2.27$ $x=2.27$ .

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How do you solve $3^{x + 4} = 7^{2 x - 1}$ $3^(x+4)=7^(2x-1)$ ?

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Explanation:

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How do you solve $3^{x + 4} = 7^{2 x - 1}$ $3^(x+4)=7^(2x-1)$ ?