How do you solve ${log}_{6} x + {log}_{6} 3 = 2$ $log_(6)x + log_(6)3 = 2$ ?

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Answer 1 · 2018-06-11T03:53:57

$x = 12$ $x=12$

We can start by subtracting ${log}_{6} (3)$ $log_6(3)$ from both sides. This gives us

${log}_{6} x = 2 - {log}_{6} (3)$ $log_6x=color(darkviolet)(2-log_6(3))$

We can apply the logarithm rule

$a = {log}_{b} (b^{a})$ $a=log_b(b^a)$

where our $a$ $a$ 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:

$6^(2-log_6(3))$ can be rewritten as

$6^(-log_6(3))*6^2$

Which simplifies to

$3^-1*color(blue)(6^2)$

Further simplifying, we get

$1/3*color(blue)(36)=36/3=color(darkviolet)(12)$

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

$log_6x=color(darkviolet)(log_6(12))$

Since the bases are the same, we now have

$x=12$

All we did was:

Hope this helps!

How do you solve log_(6)x + log_(6)3 = 2log6x+log63=2log_(6)x + log_(6)3 = 2?