How do you solve ${log}_{5} (4 x + 11) = 2$ $log_5(4x+11)=2$ ?

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Answer 1 · 2016-08-14T20:45:11

$x = 3 \frac{1}{2}$ $x = 3 1/2$

Log form and index form are interchangeable.

If ${log}_{a} b = c then \Rightarrow a^{c} = b$ $log_a b = c " then "rArr a^c = b$

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

$25 = 4 x + 11$ $25 = 4x +11$

$14 = 4 x$ $14 = 4x$

$x = 3 \frac{1}{2}$ $x = 3 1/2$

How do you solve log5(4x+11)=2log_5(4x+11)=2?