# How do you solve log_(6)x + log_(6)3 = 2?

Jun 11, 2018

$x = 12$

#### Explanation:

We can start by subtracting ${\log}_{6} \left(3\right)$ from both sides. This gives us

${\log}_{6} x = \textcolor{\mathrm{da} r k v i o \le t}{2 - {\log}_{6} \left(3\right)}$

We can apply the logarithm rule

$a = {\log}_{b} \left({b}^{a}\right)$

where our $a$ is what I have in purple above. This can be rewritten as

log_6x=color(darkviolet)(log_6(ul(6^(2-log_6(3))))

Let's simplify what I have underlined:

${6}^{2 - {\log}_{6} \left(3\right)}$ can be rewritten as

${6}^{- {\log}_{6} \left(3\right)} \cdot {6}^{2}$

Which simplifies to

${3}^{-} 1 \cdot \textcolor{b l u e}{{6}^{2}}$

Further simplifying, we get

$\frac{1}{3} \cdot \textcolor{b l u e}{36} = \frac{36}{3} = \textcolor{\mathrm{da} r k v i o \le t}{12}$

Remember, this is the underlined portion of our equation. We have

${\log}_{6} x = \textcolor{\mathrm{da} r k v i o \le t}{{\log}_{6} \left(12\right)}$

Since the bases are the same, we now have

$x = 12$

All we did was:

• Subtract to get logs on both sides
• Rewrite the expression in a form we can deal with
• Simplify the expression using many exponent properties
• Realized that the bases are the same, so they basically cancel

Hope this helps!