Question

What is the solution set for x in log4(x+5)+log4x=log4 14

Answer 1

$log_ {4} (x+5) + log_4(x) = log_4(14)$

For logs with same base :

$log_b(f(x)) + log_b(g(x)) = log_b(f(x)*g(x))$

$Log_4(x+5) + log_4(x) =log_4((x+5)x)$

$log_4((x+5) = log_4(14)$

When the logs have the same base :

$log_b(f(x)) = log_b(g(x))$

$F(x) = g(x)$

For $log_4((x+5)x) = log_4(14)$

Solve $(x+5)x = 14$

expand

$(x+5)x = x^2 + 5x$

$x^2+5x+14$

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

$x = 2 , x = - 7$

Therefore, the final solution : $x = 2$

hope this helps!