How do you solve #log_6(x+8)+log_6 7=2#?

1 Answer
Jim G.
Jul 28, 2016

#x=-20/7#

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)#
Applies to logarithms to any base.

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

From (A) #log_6(x+8)+log_6 7=log_6(7(x+8))#

From (B) #7x+56=6^2=36rArrx=-20/7#

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