How do you solve ${log}_{7} x + {log}_{7} (x + 5) = {log}_{7} (14)$ $log_7x+log_7(x+5)=log_7(14)$ ?

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Answer 1 · 2016-12-08T21:22:07

$x = - 7$ $x= -7$

$x = + 2$ $x= +2$

Notice is that everything is to logs of the same base. This means that we can get rid of the logs.

Note that $log (a) + log (b)$ $log(a)+log(b)$ is the same thing as $log (a \times b)$ $log(axxb)$

This gives us:

${log}_{7} (x (x + 5)) = {log}_{7} (14)$ $log_7(x(x+5))=log_7(14)$

$\Rightarrow x (x + 5) = 14$ $=> x(x+5)=14$

$x^{2} + 5 x - 14 = 0$ $x^2+5x-14=0$
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Note that $2 \times 7 = 14 and 7 - 2 = 5$ $2xx7=14" and "7-2=5$

$(x + 7) (x - 2) = 0$ $(x+7)(x-2)=0$

$x = - 7$ $x= -7$

$x = + 2$ $x= +2$

How do you solve log7x+log7(x+5)=log7(14)log_7x+log_7(x+5)=log_7(14)?