Jules J.
asked • 01/11/21Assessment #7 - Solve Exponential Equations (Algebra 1) This is due today
Theres is like 30 of these problems and I was hoping I could get some help to start me off
(also when i put " ^ " it means the following of it is the exponent(s) until there is a space " ")
Solve the following exponential equation for x. Be sure to show all of your work.
- 5^x = 125
- 5^x+7 = 25
- 2^2x+5 = 64
- 2^3x-8 = 16
- 3^x-3 +4 =31
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1 Expert Answer
Bradford T. answered • 01/11/21
Tutor
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Retired Engineer / Upper level math instructor
- Think of this as 5what = 125. 125 = 5•5•5 = 53. So if 5x= 53, then x = 3
- 5x+7 = 25 = 5•5 = 52, So x+7 = 2 or x = -5
- 22x+5 = 64 = 26, 2x+5 = 6, x = 1/2
- 23x-8 = 16 = 24, 3x-8 =16, x = 24/3 = 8
- 3x-3 + 4 = 31, 3x-3 = 27 = 33, So x-3 = 3, x = 6
So the pattern here is baseequation = basen. Then just equate the exponents and solve for x.
Number 5 just needs a little work to get it into that form.
The sub-lesson here is that the exponents are just logs for the given base.
Jules J.
Thank you :)
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01/12/21
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Stanton D.
Hi Jules J., The function that "unpeels" exponentiation is taking the logarithm. That can be to any number base; usually either "common logs" = namely log to the base 10, or "log(x)", or "natural logs", namely log to the base e, or "ln(x)" , are used. So for your first example, log(5^x) = log(125) ; x*log(5) = log(125) ; solves to x = 3 . In all these problems, though, there is an evident other base number on which to take a log function. In this first problem, that base is 5, and you can factor 5's out of 125 easily by eye = 5*5*5 or 5^3 . -- Cheers, --Mr. d.01/11/21