SOLUTION: Greetings I am having trouble with a log problem to solve for X in logx32=5/3 I thought I must change into exponential form and get x^5/3 =2^5 and then go into log form to

Question 633354: Greetings I am having trouble with a log problem to solve for X
in logx32=5/3
I thought I must change into exponential form and get x^5/3 =2^5
and then go into log form to get
3/5logx=5log2 and then I am lost.
please help?
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!

You made a great first step with

Not only did you get the variable out of the logarithm, but you recognized that which turns out to be a very helpful.

But it's not a good idea to go back to log form. That just puts the x back in the logarithm which is not where we want it. Instead, what we want is to change the exponent on x to a 1. If we can do that, we'll be finished since !

So how do we change an exponent? Well there are several rules for exponents that tell us: "When we do __________, we change the exponents in this way.". If we use the multiplication rule (add the exponents) to get an exponent of 1 (by multiplying by we could get the exponent of 1 on the left but then we would have on the the right side. If we use the division rule (subtract the exponents) we have a similar problem, We can get the exponent we want on the left but we end up with an x on the right side. The rule that will help us is the power of a power rule (multiply the exponents). If we raise both sides to the right power, we can get an exponent of 1 on the left and there will not be any x's on the right.

So what power to we raise to to get an exponent of 1? Knowing that we will be multiplying this exponent by 5/3 and knowing that multiplying reciprocals always results in 1, the right power to use is the reciprocal of 5/3: 3/5.

On the left, we get the 1 we wanted. On the right, after we multiply 5 and 3/5, we get:

which simplifies to:

And, as usual with problems where the variable was in the argument or base of a logarithm, we must check the solution to ensure that all arguments and bases of all logarithms remain valid (positive for both and bases cannot be a 1 either). If any argument or base ends up being invalid we must reject that "solution", even if it's the only one we found!

Use the original equation to check:

Checking x = 8:

We can quickly see that the base is positive (and not a 1) and the argument is positive. So our solution checks!
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