How do you solve $log_11 (y + 8) + log_11 4 = log_11 60$ ?

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Answer 1 · 2015-09-06T17:23:11

$y = 7$

From laws of logs, the log of the sum is the product of the logs.
So left hand side of equation reduces to

$log_11(4y + 32) = log_11(60)$

Now take the anti-log base 11 on both sides to yield:

$4y + 32 = 60$

This is now a normal linear equation in y which may be solved by normal methods to eventually get $y = 7$ .

How do you solve log_11 (y + 8) + log_11 4 = log_11 60 log_11 (y + 8) + log_11 4 = log_11 60?