# Solve `log_x 32 + log_x 16 = 9/2`

txmedteach | Certified Educator

To start this problem, you'll need to take a plan of attack! espeically in logarithm problems, the plan usually involves two things:

1) Combine all of the logarithm terms into one giant logarithm.

2) Make both sides exponents of the base of the log (i.e. if `log_a x = b` then we change that to `a^(log_a x) = a^b` which cancels out the log)

3) Clean up and solve algebraically.

This method works for many of the log problems they throw at you, and I would almost guarantee it in any equation that you could solve algebraically! I only say "almost" because there are almost always exceptions to these kinds of "rules of thumb."

So let's apply our solution method to this problem.

Step 1: Combine the logs

We'll start by figuring out how to combine the logarithms on the left-hand side.

`log_x32 + log_x16 = 9/2`

To combine these two logs, let's consider the multiplication rule for logs:

`log_a (bc) = log_a b + log_a c`

This can be easily derived from exponental relations, and it is easily applied to our problem!

`log_x 32 + log_x 16 = log_x (32*16) = log_x 512`

Now, we have combined our logs, and our equation looks like :

`log_x 512 = 9/2`

We can now move onto our second step.

Step 2: Make each term an exponent!

This may sound funny, but it will make sense after you consider how logarithms are defined. Recall, if `a^x = b` then the logarithm is defined in this way: `log_a b = x`. This means that logs and exponential functions are just inverse functions.

So, when I say make each term an exponent of the base, it is the same as dividing when you have a coefficient of x, just slightly different. Here's what this process looks like for our problem:

`log_x 512 = 9/2`

`x^(log_x512) = x^(9/2)`

Now because x^ (taking x to a power) and `log_x` are inverse functions, the `x` and `log_x` cancel! This gives us:

`512 = x^(9/2)`

Now, we move onto step 3.

Step 3: Solve the equation algebraically.

Now, we'll just have to solve this equation like we'd solve any other equation. First, though, note that `x` will have to be positive because we can rewrite the equation:

`512 = sqrt(x) ^ 9`

Assuming we're only dealing with real numbers, we can't have the square root of a negative number!

Because of this constraint on `x` we don't have to worry about factoring a polynomial or anything like that, we can just solve everything by taking both sides to the `2/9` power:

`512^(2/9) = (x^(9/2))^(2/9)`

Remember, when you take an exponent to an exponent (or a power to a power) you just multiply the exponents!

`512^(2/9) = x^(2/9 * 9/2)`

This allows us to cancel the exponent in the `x`, giving us our result.

`512^(2/9) = x`

Now, that actually turns out to be a nice number. You might know offhand that `512 = 2^9`, and because we're taking the 9th root of 512 (because 9 is in the denominator of the exponent), we can just change the number in the following way:

`512^(2/9) = root(9)(512) ^ 2 = 2^2`

Well, that's not bad! We know that `2^2 = 4` so we have our cleaned-up solution:

`x = 4`

I hope that helps!