How do you solve $log_7 3+log_7x=log_7 32$ ?

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Answer 1 · 2016-07-15T21:53:36

$x = 32/3 = 10 2/3$

If logs to the same base are being added, then the numbers were multiplied.

$log_7 3 + log_7 x = log_7 32$

$log_7(3xx x) = log_7 32$

$3x = 32" if log A = log B. then A = B"$

$x = 32/3 = 10 2/3$

Answer 2 · 2016-07-15T21:55:41

$x = 32/3$

Use the formula:

$log_a(x) = (log_b(x))/(log_b(a))$

So our equation becomes:

$(log(3))/(log(7)) + (log(x))/(log(7)) = (log(32))/(log(7))$

Multiply both sides by $log(7)$ and combine using rules of logs

$log(3x) = log(32)$

Cancelling the logs by taking exponents yields

$3x = 32$

$implies x = 32/3$

How do you solve log_7 3+log_7x=log_7 32log_7 3+log_7x=log_7 32?