How do you write $4^(x+1) = 32$ in log form?

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Answer 1 · 2016-08-25T14:27:21

$4^(x+1) = 32 " " hArr " "log_4 32 = x+1$

log form and index form are interchangeable:

$log_a b = c" " hArr" " a^c = b$

"The base stays the base and the other two swop around"

$4^(x+1) = 32 " " hArr " "log_4 32 = x+1$

How do you write 4^(x+1) = 324^(x+1) = 32 in log form?