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How to sove an equation like 5^x=80?
That's the only thing my math teacher didn't get to for our finals review. I'm having trouble finding it in my math book. I'm pretty sure I know how to solve it but I want to make sure.
5 Answers
- 8 years agoFavorite Answer
You have to make the bases equal to each other and then set the exponents into an equation and solve. Ex: 3^x=9....: 3^x=3^2....: x=2
- 8 years ago
Get the logarithm of both sides: ln 5^x=ln 80
Then apply the property of logarithm: xln 5=ln 80
Divide both sides by ln 5 to solve for x: x=ln 80/ln 5
- 8 years ago
This is a very very very very simple question
5^x=80
x=80/5
Therefor x=16
Source(s): My knowledge - Anonymous8 years ago
To bring down the exponent "x", take logarithm on both sides. Logarithm has a property:
log A^c = c log A
log(5^x)=log 80
xlog5=log80
x= (log80)/log(5)
x=2.72
Source(s): Joseph's 487 questions to Algebra 2 - How do you think about the answers? You can sign in to vote the answer.
- 8 years ago
5^x = 80
taking logarithm on both sides
log(5^x) = log(80)
x*log(5) = log(80)
=> x = log(80)/log(5)
