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

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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:

Substitute for the 4:

Using the rules of exponents:

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 :
log2(x+1) - log2(x^(1/2)) = 1
log2(x+1) - log2() = 1
log2
Rewriting in exponential form:

Multiplying both sides by

Squaring both sides:

Subtracting 4x from both sides

Factoring:

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.

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