How do you solve #logx-log2=1#?

1 Answer
Aug 10, 2016

Answer:

x = 20

Explanation:

Using the following #color(blue)"laws of logarithms"#

#color(orange)"Reminder"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(logx-logy=log(x/y))color(white)(a/a)|)))........ (A)#
This applies to logarithms to any base.

#color(red)(|bar(ul(color(white)(a/a)color(black)(log_b a=nhArra=b^n)color(white)(a/a)|)))........ (B)#

#"Using (A) " logx-log2=log(x/2)#

A logarithm expressed as log x , usually indicates that the base is 10.

#"Using (B) " log_(10)(x/2)=1rArrx/2=10^1=10#

Thus #x/2=10rArrx=20#