SOLUTION: I am solving a 7th grade math text book, and I came across this one: a^x = b^y = c^z, and a� = b�c, then (3/x) - (2/y) = ? I tried: --> c = (a�/b�) --> a = cube root(b�c)

Question 825465: I am solving a 7th grade math text book, and I came across this one:
a^x = b^y = c^z, and a� = b�c, then (3/x) - (2/y) = ?
I tried:
--> c = (a�/b�)
--> a = cube root(b�c)
--> b = √(a�/c),
but couldn't arrive at a conclusion; and also thought there should be a simpler way of solving this.
Please help.
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
If this really came out of a 7th grade book, then there must be a much simpler way to solve this than the one I am about to show you. I just can't see anything simpler than what you are about to see. If you believe that there is a simpler solution, then please re-post your question and specify "without using logarithms" so you don't get a solution like below.

As I figured out this solution, my thought process was...

To find (3/x) - (2/y) we will need to find an expression for 3/x and for 2/y
To find expressions for 3/x and 2/y we will need an expression for x and for y. (A simpler expression, if one exists, probably does not make this assumption.)
To find expressions for x and for y we will need to use logarithms since the only place we can find them at the start is in exponents (and logarithms are a tool which can be used to solve for an exponent). (Logarithms are almost never taught in 7th grade. This is why I assume there must be a different, simpler solution.)

To start, I will introduce another, temporary variable (which we'll name "q") and say the q is also equal to all those powers of a, b and c:

Now we will solve for x, y and z. First we will use

to solve for x. Finding the log of each side:

Next we use a property of logarithms, , to "move" the exponent out in front:

Dividing both sides by log(a):

Now we will find an expression for 3/x. Taking the reciprocal of each side:

Multiplying each side by 3:

Now we will use
to get an expression for y first, then an expression for 2/y. The steps are similar to the ones we used above so I will not explain them again:

Now we will use to get an expression for z:

Now let's look at what we have for 3/x - 2/y. Substituting in the expressions we found above we should get:

These already have the same denominator so we can subtract:

Not very interesting yet. But if we use that property for logarithms, in the other direction, we can move the coefficients of the logs into the arguments:

Still not very interesting. But if we use another property of logarithms, , to combine these logs:

Starting to look interesting... From , we can get . Substituting this in we get:

This looks vaguely familiar. Looking back we can see that

The left side of this is the reciprocal of what we currently have for 3/x - 2/y! So:

P.S. If you find a simpler, logarithm-less solution to this, please tell me what it is via a "thank you" for this solution.
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