How do you solve $7^{x} + 2 = 410$ $7^x + 2 = 410$ ?

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Answer 1 · 2016-05-20T09:27:34

3.09

$7^{x} + 2 = 410$ $7^x + 2 =410$

$7^{x} = 408$ $7^x = 408$

Take the natural log of both sides,we get,

$ln (7^{x}) = ln (408)$ $ln(7^x) = ln (408)$

Using logarithm properties, we can bring the x exponent outside the ln

$x ln (7) = ln (408)$ $xln(7) = ln (408)$

$x = \frac{ln (408)}{ln (7)}$ $x=ln(408) / ln(7)$

$= (6.0112)$ $=(6.0112)$ / $(1.945)$ $(1.945)$

$3.09$ $3.09$

How do you solve 7x+2=4107^x + 2 = 410?