Question 591606: log4(x+2)-log4(x-1)=1
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! log(4,x+2) - log(4,x-1) = 1
since, in general, log(x) - log(y) = log(x/y), this equation becomes:
log(4,(x+2)/(x-1)) = 1
since, in general, log(b,x) = y if and only if b^y = x, this equation becomes:
4^1 = (x+2)/(x-1) which is the same as:
(x+2)/(x-1) = 4
multiply both sides of this equation by (x-1) to get:
(x+2) = 4*(x-1)
simplify to get:
x+2 = 4x-4
subtract x from both sides of this equation and add 4 to both sides of this equation to get:
6 = 3x which is the same as:
3x = 6
divide both sides of this equation by 3 to get:
x = 2
replace x with 2 in your original equation.
log(4,x+2) - log(4,x-1) = 1 becomes:
log(4,(4) - log(4,1) = 1
this is true if and only if 4^1 = 4
since 4^1 is equal to 4, this confirms the solution of x = 2 is good.
you can also confirm it by solving it using the LOG function of your calculator.
to do that, you need to convert the equation as follows:
log(4,(4) - log(4,1) = 1 becomes:
LOG(4)/LOG(4) - LOG(1)/LOG(4) = 1
use the log function of your calculator to solve to get:
you get:
1 - 0 = 1 which becomes 1 = 1 which is true, confirming x = 2 is good.
LOG(4) / LOG(4) is equivalent to log(10,4) / log(10,4)
LOG(1) / LOG(1) is equivalent to log(10,1) / log(10,4)
since, in general, x/x = 1, LOG(4)/LOG(4) = 1
since log(10,1) = y if and only if 10^y = 1, this occurs only when y = 0 which makes the expression log(10,1) equivalent to 0 which is confirmed when you use your calculator to solve for LOG(1). you get 0 as the answer.
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