How do I solve this? thank you

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1 Answer
Jim G.
Dec 10, 2016

#x=2#

Explanation:

Hi Danny. I am assuming you mean the following.

#log_x64=6#

Use the #color(blue)"law of logarithms"#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(log_bx=nhArrx=b^n)color(white)(2/2)|)))#

#"Here " x=64,b=x" and " n=6#

#rArr64=x^6#

#"Now " 64=2^6larr" worth knowing powers of 2"#

#"Thus "2^6=x^6rArrx=2#

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