How do you solve the equation #2 log4(x + 7)-log4(16) = 2#?

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Answer 1 · 2016-08-11T21:36:17

# x= 9#

When working with a log equation, the term must be all logs or all numbers.
Change the RHS to a log term as well.

#color(red)(2)log_4 (x+7) - log_4 16 = log_4 16 " "2 = log_4 16#

#log_4 (x+7)^color(red)(2) - log_4 16 = log_4 16#

"If logs are being subtracted, the numbers are being divided"

#log_4 ((x+7)^2/16)= log_4 16#

#(x+7)^2/16 = 16" if log A = log B, A = B"#

#(x+7)^2 = 256" cross multiply"#

#x+7 = 16" only pos root is valid"#

# x= 9#