Question 1199920: 5^(x+1)=10
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! start with 5^(x+1) = 10
take the log of both sides of the equation to get:
log(5^(x+1)) = log(10)
this becomes:
(x+1) * log(5) = log(10)
divide both sides of the equation by log(5) to get:
x + 1 = log(10)/log(5)
solve for x to get:
x = log(10)/log(5) - 1 = .4306765581.
that's your answer.
replace x with that in the original equation to get:
5^(.4306765581+1) = 10 which results in 10 = 10, confirming the value of x is correct.
you could also use the basic logarithm property to solve this as well.
that property is:
logb(x) = y if and only if b^y = x
conversely, b^y = x if and only if logb(x) = y
working with the converse equation of b^y = x if and only if logb(x) = y, you get:
when b = 5 and y = x+1 and x = 10, you get:
b^y = x if and only if logb(x) = y becomes:
5^(x+1) = 10 if and only if log5(10) = x+1
solve for x to get:
x = log5(10) - 1
now, log5(10) = log(10) / log(5) by the log base conversion formulas, so you get:
x = log(10) / log(5) - 1 = .4306765581.
your original equation becomes:
5^(.4306765581 + 1) = 10 which becomes:
10 = 10, confirming the value of x is correct.
either way comes out to the same answer, which is:
x = .4306765581.
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