How do you solve $2^{x} = 5^{x - 2}$ $2^x = 5^(x - 2)$ ?

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Answer 1 · 2016-10-29T22:03:39

How you do it is explained below.

Take the natural logarithm of both sides:

$ln (2^{x}) = ln (5^{x - 2})$ $ln(2^x) = ln(5^(x - 2))$

Use the identity $ln (a^{b}) = ln (a) b$ $ln(a^b) = ln(a)b$

$ln (2) x = ln (5) (x - 2)$ $ln(2)x = ln(5)(x - 2)$

Distribute ln(5):

$ln (2) x = ln (5) x - 2 ln (5)$ $ln(2)x = ln(5)x - 2ln(5)$

Subtract ln(5)x from both sides:

$(ln (2) - ln (5)) x = - 2 ln (5)$ $(ln(2)-ln(5))x = -2ln(5)$

Multiply both sides by -1:

$(ln (5) - ln (2)) x = 2 ln (5)$ $(ln(5)-ln(2))x = 2ln(5)$

Divide both sides by $(ln (5) - ln (2))$ $(ln(5)-ln(2))$ :

$x = \frac{2 ln (5)}{ln (5) - ln (2)}$ $x = (2ln(5))/(ln(5)-ln(2))$

$x \approx 3.5$ $x ~~ 3.5$

How do you solve 2^x = 5^(x - 2)2x=5x−22^x = 5^(x - 2)?