How do you solve ${log}_{7} (x + 3) - 2 = {log}_{7} 3$ $log_7 (x+3) - 2 = log_7 3$ ?

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Answer 1 · 2016-05-11T05:31:22

$x = \frac{40}{3}$ $x=40/3$

Use log m + log m = log (mn).

Here, ${log}_{7} (x + 3) + {log}_{7} 3 = {log}_{7} (3 (x + 3)) = 2$ $log_7(x+3)+log_7 3= log_7(3(x+3))=2$ .

Use that if $c = {log}_{b} a, a = b^{c}$ $c=log_b a, a=b^c$ .

Accordingly, $3 (x + 3) = 7^{2} = 49$ $3(x+3)=7^2=49$ .

So, $x = \frac{40}{3}$ $x = 40/3$ .

How do you solve log7(x+3)−2=log73log_7 (x+3) - 2 = log_7 3?