How do you solve $25^{x} - 8 \cdot 5^{x} = - 16$ $25^ { x } - 8\cdot 5^ { x } = - 16$ ?

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Answer 1 · 2018-06-22T07:00:11

$x = {log}_{5} (2)$ $x=log_5(2)$

${(5^{2})}^{x} - 8 \cdot 5^{x} = - 16$ $(5^2)^x-8 *5^x=-16$
$t^{2} - 8 t = - 16$ $t^2-8t=-16$
$t = 4$ $t=4$
$5^{x} = 4$ $5^x=4$

Answer 2 · 2018-06-22T07:01:01

$x = \frac{log 4}{log 5} = 0.8614$ $color(maroon)(x = log 4 / log 5 = 0.8614$

$25^{x} - 8 \cdot 5^{x} + 16 = 0$ $25^x - 8 * 5^x + 16 = 0$

Let $5^{x} = a$ $5^x = a$

$\Rightarrow a^{2} - 8 a + 4^{2} = 0$ $=> a^2 - 8a + 4^2 = 0$

${(a - 4)}^{2} = 0$ $(a-4)^2 = 0$

$:. a = 4$

$a = 5^x = 4$

$x log 5 = log 4$

$x = log(4) / log(5) = 0.8614$

How do you solve 25^ { x } - 8\cdot 5^ { x } = - 1625x−8⋅5x=−1625^ { x } - 8\cdot 5^ { x } = - 16?