SOLUTION: Solve for x log (9 ^ log base3 x+1) + log (2x+3/x+1) = log base4 8

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Question 271622: Solve for x

log (9 ^ log base3 x+1) + log (2x+3/x+1) = log base4 8
Answer by jsmallt9(3758)   (Show Source): You can put this solution on YOUR website!
I'm assuming the equation is

I used base 4 logarithms on the left side because without them the problem is extremely difficult. I also had to guess the numerator and denominator in the argument of the second logarithm. If this is wrong, repost your question. For complex expressions like those in your equation:
For example, here is the text I used for the equation above (without the braces so you can see the text itself:
log(4, (9 ^ log(3, (x+1)))) + log(4, ((2x+3)/(x+1))) = log(4, (8))

If my equation is the correct one, here is a way to solve it. (If my equation is wrong, you may still learn something from this.) First, we'll work on simplifying the first logarithm. The argument is 9 to a base 3 logarithm power. Since 9 is a power of 3, we can simplify the argument by rewriting the argument using 3 squared instead of 9:

Using the rule for exponents for a power of a power (i.e. multiply the exponents) we get:

Now we can use the property of logarithms, , we can move the 2 in front of the log into the argument as an exponent:

Now the argument of the first logarithm is 3 to a base 3 logarithm power. If we understand what logarithms represent, we can simplify this argument. represents the exponent for 3 that results in . And this expression is being used as an exponent for 3! So !

The first logarithm is now greatly simplified. Our next goal is to combine the two logarithms on the left. This will give us an equation of the form:
log(expression) = log(other-expression)
which can be solved. These logarithms are not like terms so they cannot be added. But there is a property of logarithms, , which allows us to combine two logarithms with a "+" between them into a single logarithm. Using this property on the two logarithms in the equation we get:

one of the two (x+1)'s in the numerator cancels the (x+1) in the denominator leaving:

Now the equation says that the base 4 logarithm of (x+1)(2x+3) equals the base 4 logarithm of 8. If this is true then (x+1)(2x+3) must be 8:

The logarithms are now gone!. This equation can now be solved. Multiplying out the left side we get:

Subtracting 8 from each side we get:

Factoring the left side we get:

From the Zero Product Property we know that one of the factors must be zero. So:
or
Solving these we get:
or

When solving logarithmic equations you should always check your answers. Even if we've done everything correctly so far, we may have solutions that do not fit the original equation.

Checking x = 1/2:

which simplifies to:


At this point we can see that all the logarithms have positive arguments. Logarithms of all bases must always have positive arguments. The primary reason we check our answers in problems like this is to make sure the solutions make all the arguments to logarithms positive. So x = 1/2 looks like it is going to work. If we finish the check we'll find that x = 1/2 is a solution.

Checking x = -5:

which simplifies to:


Here we can see that one of the arguments is negative. This cannot happen. We must reject x = -5 because it makes an argument to a logarithm negative.

So the only solution to your equation is x = 1/2.

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