How do you solve #logx-log2=1#?
1 Answer
Aug 10, 2016
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#
