# How do you solve 2 - log_3sqrt(x² +17) = 0 ?

Jun 2, 2016

$x = \pm 8$

#### Explanation:

$2 = {\log}_{3} \left(\sqrt{{x}^{2} + 17}\right)$

${3}^{2} = \sqrt{{x}^{2} + 17}$

$9 = \sqrt{{x}^{2} + 17}$

${\left(9\right)}^{2} = {\left(\sqrt{{x}^{2} + 17}\right)}^{2}$

$81 = {x}^{2} + 17$

$81 - 17 = {x}^{2}$

$64 = {x}^{2}$

$\pm 8 = x$

Slightly alternative approach:

$2 = {\log}_{3} \left(\sqrt{{x}^{2} + 17}\right)$

Use the rule $\sqrt{x} = {x}^{\frac{1}{2}}$

$2 = {\log}_{3} {\left({x}^{2} + 17\right)}^{\frac{1}{2}}$

Now use the rule $\log {a}^{n} = n \log a$

$2 = \frac{1}{2} {\log}_{3} \left({x}^{2} + 17\right)$

$\frac{2}{\frac{1}{2}} = {\log}_{3} \left({x}^{2} + 17\right)$

$4 = {\log}_{3} \left({x}^{2} + 17\right)$

${3}^{4} = {x}^{2} + 17$

$81 - 17 = {x}^{2}$

$64 = {x}^{2}$

$x = \pm 8$

Hopefully this helps!

Jun 2, 2016

simplify using the properties of logarithms to find $x = \pm 8$

#### Explanation:

Lets start by putting the term with the $x$ on one side of the equation and the constant on the other:

${\log}_{3} \sqrt{{x}^{2} + 17} = 2$

The argument of the ${\log}_{3}$ can also be rewritten as a power

${\log}_{3} \left[{\left({x}^{2} + 17\right)}^{\frac{1}{2}}\right] = 2$

We can then use the property of the logarithm that

$\log {a}^{b} = b \log a$

to rewrite our equation as

$\frac{1}{2} {\log}_{3} \left({x}^{2} + 17\right) = 2$

then multiplying both sides by $2$ we get

${\log}_{3} \left({x}^{2} + 17\right) = 4$

To get rid of the ${\log}_{3}$ we can put both sides of the equation to the power of $3$, remembering that ${a}^{{\log}_{a} \left(b\right)} = b$, which gives us

$\left({x}^{2} + 17\right) = {3}^{4} = 81$

Subtracting $17$ from both sides we get

${x}^{2} = 64$

and taking the square root we get

$x = \pm 8$