How do I solve this equation?

#log_2(5) + log_2(2x-1) = 5#

2 Answers
May 24, 2018

#x=37/10#

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

#x = 3.7#

Explanation:

We have the rule

#log(ab)=log(a)+log(b)#

Using it "backwards", we have

#log_2(5)+log_2(2x-1) = log_2(5(2x-1)) = log_2(10x-5)#

The equation becomes

#log_2(10x-5) = 5#

Consider both sides as exponents of two:

#2^{log_2(10x-5)} = 2^5#

#therefore 10x-5 = 32#

#therefore 10x = 37#

#therefore x = 37/10 = 3.7#