Question 825465
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.<br>
As I figured out this solution, my thought process was...<ul><li>To find (3/x) - (2/y) we will need to find an expression for 3/x and for 2/y</li><li>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.)</li><li>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.)</li></ul>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:
{{{q = a^x = b^y = c^z}}}
Now we will solve for x, y and z. First we will use
{{{q = a^x}}}
to solve for x. Finding the log of each side:
{{{log((q)) = log((a^x))}}}
Next we use a property of logarithms, {{{log(m, (p^n)) = n*log(m, (p))}}}, to "move" the exponent out in front:
{{{log((q)) = x*log((a))}}}
Dividing both sides by log(a):
{{{log((q))/log((a)) = x}}}
Now we will find an expression for 3/x. Taking the reciprocal of each side:
{{{log((a))/log((q)) = 1/x}}}
Multiplying each side by 3:
{{{3log((a))/log((q)) = 3/x}}}<br>
Now we will use
{{{q = b^y}}} 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:
{{{q = b^y}}}
{{{log((q)) = log((b^y))}}}
{{{log((q)) = y*log((b))}}}
{{{log((q))/log((b)) = y}}}
{{{log((b))/log((q)) = 1/y}}}
{{{2log((b))/log((q)) = 2/y}}}<br>
Now we will use {{{q = c^z}}} to get an expression for z:
{{{q = c^z}}}
{{{log((q)) = log((c^z))}}}
{{{log((q)) = z*log((c))}}}
{{{log((q))/log((c)) = z}}}<br>
Now let's look at what we have for 3/x - 2/y. Substituting in the expressions we found above we should get:
{{{3log((a))/log((q)) - 2log((b))/log((q))}}}
These already have the same denominator so we can subtract:
{{{(3log((a))- 2log((b)))/log((q)) }}}
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:
{{{(log((a^3))- log((b^2)))/log((q)) }}}
Still not very interesting. But if we use another property of logarithms, {{{log(m, (p)) - log(m, (v)) = log(m, (p/v))}}}, to combine these logs:
{{{log((a^3/b^2))/log((q)) }}}
Starting to look interesting... From {{{a^3 = b^2 *c}}}, we can get {{{a^3/b^2 = c}}}. Substituting this in we get:
{{{log((c))/log((q)) }}}
This looks vaguely familiar. Looking back we can see that
{{{log((q))/log((c)) = z}}}
The left side of this is the reciprocal of what we currently have for 3/x - 2/y! So:
{{{3/x-2/y=1/z}}}<br>
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.