How do you solve $log_5(4x+11)=2$ ?

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Answer 1 · 2016-08-14T20:45:11

$x = 3 1/2$

Log form and index form are interchangeable.

If $log_a b = c " then "rArr a^c = b$

We have: $log_5 (4x+11) = 2 " then " 5^2 = 4x +11$

$25 = 4x +11$

$14 = 4x$

$x = 3 1/2$

How do you solve log_5(4x+11)=2log_5(4x+11)=2?