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

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Answer 1 · 2016-12-08T21:22:07

$x= -7$

$x= +2$

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

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

This gives us:

$log_7(x(x+5))=log_7(14)$

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

$x^2+5x-14=0$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Note that $2xx7=14" and "7-2=5$

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

$x= -7$

$x= +2$

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