How do you solve #log_2 10 + log_2 x = 6#?
1 Answer
Aug 11, 2016
Explanation:
Using the
#color(blue)"laws of logarithms"#
#color(orange)"Reminder"#
#color(red)(|bar(ul(color(white)(a/a)color(black)(logx+logy=log(xy))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)
#log_2 10+log_2 x=log_2(10x)# Using (B)
#log_2(10x)=6hArr10x=2^6=64#
#rArr10x=64rArrx=32/5#