How do you find ${log}_{6} (x + 4) = {log}_{6} 7$ $\log _ { 6} ( x + 4) = \log _ { 6} 7$ ?

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Answer 1 · 2018-03-02T07:24:13

The answer is $x = 3$ $x=3$ .

Use the definition of a logarithm:

$color(white)=>log_color(red)a(color(green)b)=color(blue)xqquad<=>qquadcolor(red)a^color(blue)x=color(green)b$

In this case, $a$ is $6$ , $b$ is $x+4$ , and $x$ is $log_6(7)$ . It sounds kind of confusing, but I'll use colors so it's easier to see:

$color(white)=>log_color(red)6(color(green)(x+4))=color(blue)(log_6(7))$

$=>color(red)6^color(blue)(log_6(7))=color(green)(x+4)$

Now, the $6$ and $log_6$ cancel out:

$color(white)=>color(red)cancel(color(black)6)^(color(red)cancel(color(black)(log_6))(7))=x+4$

$=>7=x+4$

$color(white)=>3=x$

The answer is $x=3$ .

How do you find \log _ { 6} ( x + 4) = \log _ { 6} 7log6(x+4)=log67\log _ { 6} ( x + 4) = \log _ { 6} 7?