How do I solve this equation?
#log_2(5) + log_2(2x-1) = 5#
2 Answers
May 24, 2018
Explanation:
#"using the "color(blue)"laws of logarithms"#
#•color(white)(x)logx+logy=log(xy)#
#•color(white)(x)log_b x=nhArrx=b^n#
#log_2(5(2x-1))=5#
#log_2(10x-5)=5#
#10x-5=2^5=32#
#10x=32+5=37#
#rArrx=37/10#
May 24, 2018
Explanation:
We have the rule
Using it "backwards", we have
The equation becomes
Consider both sides as exponents of two: