How do you solve #\log _ { 4} ( 4x - 9) = 2#?

1 Answer
Jim G.
Jul 1, 2017

#x=25/4#

Explanation:

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

#•color(white)(x)log_bx=nhArrx=b^n#

#log_4(4x-9)=2#

#rArr4x-9=4^2=16#

#rArr4x=25rArrx=25/4#

#color(blue)"As a check"#

#log_4(4xx25/4-9)=log_4 16=2#

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