# How do you evaluate log_a x = 3/2log_a 9 + log_a 2?

Feb 27, 2016

We will use the following log properties: $a \log n = \log {n}^{a} \mathmr{and} {\log}_{a} n + {\log}_{a} m = {\log}_{a} \left(n \times m\right)$

#### Explanation:

${\log}_{a} x = {\log}_{a} {9}^{\frac{3}{2}} + {\log}_{a} 2$

${\log}_{a} x = {\log}_{a} \sqrt{{9}^{3}} + {\log}_{a} 2$

${\log}_{a} x = {\log}_{a} 27 + {\log}_{a} 2$

${\log}_{a} x = {\log}_{a} \left(27 \times 2\right)$

${\log}_{a} x = {\log}_{a} 54 \to x = 54$

Practice exercises

1. Solve for x in the following equation. Round your answers to the nearest hundredth: ${\log}_{n} 2 x = \frac{1}{3} {\log}_{n} 64 - 2 \log 3$

2. A commonly used logarithm rule is if $a = b \to \log a = \log b$. Solve for x in ${2}^{2 x - 3} = {3}^{3 x}$

Good luck!

Feb 27, 2016

x = 54

#### Explanation:

using the $\textcolor{b l u e}{\text{ laws of logarithms }}$

• logx + logy = logxy

• logx - logy = log(x/y)

• logx^n = nlogx hArr nlogx = logx^n

• "If " log_b x = log_b y rArr x = y

$\Rightarrow {\log}_{a} x = {\log}_{a} {9}^{\frac{3}{2}} + {\log}_{a} 2$

[ now ${9}^{\frac{3}{2}} = {\left(\sqrt{9}\right)}^{3} = {3}^{3} = 27$]

$\Rightarrow {\log}_{a} x = {\log}_{a} \left(27 \times 2\right) = {\log}_{a} 54$

hence x = 54

Feb 27, 2016

$x = 54$

#### Explanation:

We are given ${\log}_{a} x = \frac{3}{2} {\log}_{a} 9 + {\log}_{a} 2$. The first thing I want to do is combine like- terms. To do that, I first must change $\frac{3}{2} {\log}_{a} 9$ in to the same form as ${\log}_{a} 2$. That means that I just take the coefficient, $\frac{3}{2}$, and I raise $9$ to that power, so that $\frac{3}{2} {\log}_{a} 9$ becomes ${\log}_{a} {9}^{\frac{3}{2}}$. I can do thanks to the third rule of logs. This says that "Logarithmic Rule 3: log_b(m^n) = n · log_b(m)."

Anyways, I now have my equation as ${\log}_{a} x = {\log}_{a} {9}^{\frac{3}{2}} + {\log}_{a} 2$, which can be simplified to ${\log}_{a} x = {\log}_{a} 27 + {\log}_{a} 2$. Now we can combine like-terms. Something you should know is that logarithms are a little backward. For example, we combine ${\log}_{a} 27 + {\log}_{a} 2$ by multiplying the $27$ and the $2$, like this ${\log}_{a} \left(27 \cdot 2\right)$. This is according to the first law of $\log$s.

This now gives us ${\log}_{a} x = {\log}_{a} 54$. From here, I will rewrite this $\log$ into exponetial form. The format for that is as follows: ${\textcolor{b l u e}{b}}^{\textcolor{red}{x}} = \textcolor{g r e e n}{y}$ becomes ${\log}_{\textcolor{b l u e}{b}} \textcolor{g r e e n}{y} = \textcolor{red}{x}$ and vice versa.

Applying that logic, I take ${\log}_{\textcolor{b l u e}{a}} \textcolor{red}{x} = \textcolor{g r e e n}{{\log}_{a} 54}$ and rewrite it as ${\textcolor{b l u e}{\cancel{a}}}^{\textcolor{g r e e n}{\cancel{{\log}_{a}} 54}} = \textcolor{red}{x}$. Now, because $a$ and ${\log}_{a}$ are inverses of each other, they cancel out, which brings down $\textcolor{g r e e n}{54}$, leaving us with $\textcolor{g r e e n}{54} = \textcolor{red}{x}$. There we go, we are done! Nice job.