How do you solve #log_5 9 - log_5 (x-5)=log_5 45#?

1 Answer
Jim G.
Jul 30, 2016

#x=26/5#

Explanation:

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

#color(orange)"Reminder"#

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

#color(red)(|bar(ul(color(white)(a/a)color(black)(log_b x=log_b yrArrx=y)color(white)(a/a)|)))........ (B)#

Using (A)

#log_5 9-log_5(x-5)=log_5(9/(x-5))#

Using (B)

#log_5(9/(x-5))=log_5 45rArr9/(x-5)=45#

solve # 45(x-5)=9rArrx=26/5#

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