How do you solve $3^{x} = 3 \sqrt{2}$ $3^x=3sqrt2$ ?

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Answer 1 · 2016-11-04T01:05:19

$x = \frac{ln 18}{ln 9}$ $x = ln18/ln9$

Take the natural logarithm of both sides.

$ln (3^{x}) = ln (3 \sqrt{2})$ $ln(3^x) = ln(3sqrt(2))$

Use the logarithm rule ${ln a}^{n} = n ln a$ $lna^n = nlna$ .

$x ln 3 = ln (3 \sqrt{2})$ $xln3 = ln(3sqrt(2))$

$x ln 3 = ln (\sqrt{18})$ $xln3 = ln(sqrt(18))$

Use the same logarithm rule as above.

$x = \frac{{ln (18)}^{\frac{1}{2}}}{ln 3}$ $x = ln(18)^(1/2)/ln3$

$x = \frac{\frac{1}{2} ln 18}{ln 3}$ $x= (1/2ln18)/ln3$

$x = \frac{ln 18}{2 ln 3}$ $x = ln18/(2ln3)$

$x = \frac{ln 18}{ln 9}$ $x = ln18/ln9$

Hopefully this helps!

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