How do you solve $3 \cdot 5^{x - 1} + 5^{x} = 0.32$ $3 * 5^(x-1) + 5^x = 0.32$ ?

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How do you solve $3 \cdot 5^{x - 1} + 5^{x} = 0.32$ $3 * 5^(x-1) + 5^x = 0.32$ ?

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Answer 1 · 2016-05-15T21:25:55

$x = - 1$ $x = -1$

Explanation:

The equation can be written as $\frac{3}{5} 5^{x} + 5^{x} = (\frac{3}{5} + 1) 5^{x} = 0.32$ $3/5 5^x + 5^x = (3/5+1)5^x =0.32$
arriving at $5^{x} = 0.2 = 5^{- 1}$ $5^x = 0.2 = 5^(-1)$ then equating exponents $x = - 1$ $x = -1$

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How do you solve $3 \cdot 5^{x - 1} + 5^{x} = 0.32$ $3 * 5^(x-1) + 5^x = 0.32$ ?

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Explanation:

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How do you solve 3 * 5^(x-1) + 5^x = 0.323⋅5x−1+5x=0.323 * 5^(x-1) + 5^x = 0.32?

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How do you solve $3 \cdot 5^{x - 1} + 5^{x} = 0.32$ $3 * 5^(x-1) + 5^x = 0.32$ ?