How do you solve #\log _ { 4} ( m - 3) + \log _ { 4} ( m + 3) = 2#?

2 Answers
Mar 31, 2018

#m=5#

Explanation:

#"using the "color(blue)"laws of logarithms"#

#•color(white)(x)logx+logyhArrlog(xy);x,y>0#

#•color(white)(x)log_b x=nhArrx=b^n#

#rArrlog_4(x-3)(x+3)=2#

#rArrx^2-9=4^2=16#

#rArrx^2=16+9=25#

#rArrx=+-5#

#"but "m-3>0" and "m+3>0#

#rArrx=5" is the solution ",x=-5" is invalid"#

Mar 31, 2018

#m=5#

Explanation:

All the terms need to be in the same form. Either all logs, or all numbers.

Change #2# into a log of base #4#

#log_4 x = 2 hArr x = 4^2 = 16#

#log_4 (m-3) +log_4 (m+3) = log_4 16#

Recall a law of logs: # color(blue)(log (axxb) = log a+ log b)#

Multiplying numbers is the same as adding their logs.

Adding logs is the same as multiplying the numbers.

#log_4 ((m-3)(m+3)) = log_4 16#

Recall another log law. If #color(blue)( log a = log b, " then " a = b)#

#(m-3)(m+3) = 16#

#m^2 - 9 =16#

#m^2 = 25#

#m = +-sqrt25#

#m = +5 or m = -5#

However, logs are undefined for negative values, so #-5# is not a valid solution.

#m =5#