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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!
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