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#