SOLUTION: if you cant find the book then the question is log2 (x+1)-log4 x=1 i am having trouble solving for x the 2 and the 4 are the bases of the logs

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: if you cant find the book then the question is log2 (x+1)-log4 x=1 i am having trouble solving for x the 2 and the 4 are the bases of the logs       Log On


   

Question 201770: if you cant find the book then the question is
log2 (x+1)-log4 x=1
i am having trouble solving for x
the 2 and the 4 are the bases of the logs


Found 2 solutions by jsmallt9, RAY100:
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log2(x+1)- log4(x) = 1

First we need to get the bases the same. There is a base-changing formula but I can't recall it at the moment. But since 4 is a power of two we can convert the bases without the formula.

Let y = log4(x)
Rewrite this in exponential form:
4%5Ey+=+x
Substitute 2%5E2 for the 4:
%282%5E2%29%5Ey+=+x
Using the rules of exponents:
2%5E%282y%29+=+x
Find the log2 of both sides:
2y = log2(x)
Multiply both sides by 1/2:
y = (1/2)*log2(x)
But y = log4(x)
So log4(x) = (1/2)*log2(x)
Substituting into our equation:
log2(x+1) - (1/2)*log2(x) = 1
Using properties of logarithms and the fact that x%5E%281%2F2%29+=+sqrt%28x%29:
log2(x+1) - log2(x^(1/2)) = 1
log2(x+1) - log2(sqrt%28x%29) = 1
log2%28%28x%2B1%29%2F%28sqrt%28x%29%29%29+=+1
Rewriting in exponential form:
%28x%2B1%29%2Fsqrt%28x%29+=+2%5E1
Multiplying both sides by sqrt%28x%29
x+%2B+1+=+2%2Asqrt%28x%29
Squaring both sides:
x%5E2+%2B2x+%2B+1+=+4x
Subtracting 4x from both sides
x%5E2+-+2x+%2B+1+=+0
Factoring:
%28x+-+1%29%5E2+=+0
The solution to this will be the x-values that make
x - 1 = 0
Adding 1 to both sides gives
x = 1
Checking out answer (which is important because we must reject x-values that might make the arguments of log functions <=0 and because squaring both sides can introduce false solutions):
log2(1+1) - log4(1) = 1
Since log2(2) = 1 and log4(1) is 0, out answer checks.

Answer by RAY100(1637) About Me  (Show Source):
You can put this solution on YOUR website!
log(2) {x+1} - log(4) { x) =1
.
{log(x+1)/log 2} -{ log(x)/log4} =1
.
by observation,,,if x=1
.
{log 2 /log2} - {log(1)/log4} =1
.
1-0 =1
.
1=1
.
therefore x=1