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

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Answer 1 · 2018-06-22T10:00:05

$x = 9.5414$ $x = 9.5414$

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

$\Rightarrow (x + 4) log 2 = (x - 1) log 3$ $=>(x + 4) log 2 = (x-1) log 3$

$\Rightarrow (x + 4) \cdot 0.301 = (x - 1) \cdot 0.4772$ $=> (x+4) * 0.301 = (x-1) * 0.4772$

$x (0.4772 - 0.301) = (4 \cdot 0.301 + 0.4772)$ $x (0.4772 - 0.301) = (4 * 0.301 + 0.4772)$

$0.1762 x = 1.6812$ $0.1762 x = 1.6812$

$x = \frac{1.6812}{0.1762} = 9.5414$ $x = 1.6812 / 0.1762 = 9.5414$

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