How do you solve #log_2 10 + log_2 x = 6#?

1 Answer
Jim G.
Aug 11, 2016

#x=32/5#

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#

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