document.write( "Question 825465: I am solving a 7th grade math text book, and I came across this one:\r
\n" ); document.write( "\n" ); document.write( "a^x = b^y = c^z, and a³ = b²c, then (3/x) - (2/y) = ?\r
\n" ); document.write( "\n" ); document.write( "I tried:
\n" ); document.write( "--> c = (a³/b²)
\n" ); document.write( "--> a = cube root(b²c)
\n" ); document.write( "--> b = √(a³/c),
\n" ); document.write( "but couldn't arrive at a conclusion; and also thought there should be a simpler way of solving this.\r
\n" ); document.write( "\n" ); document.write( "Please help. \n" ); document.write( "
Algebra.Com's Answer #497377 by jsmallt9(3758)\"\" \"About 
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.

\n" ); document.write( "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:
\n" ); document.write( "\"q+=+a%5Ex+=+b%5Ey+=+c%5Ez\"
\n" ); document.write( "Now we will solve for x, y and z. First we will use
\n" ); document.write( "\"q+=+a%5Ex\"
\n" ); document.write( "to solve for x. Finding the log of each side:
\n" ); document.write( "\"log%28%28q%29%29+=+log%28%28a%5Ex%29%29\"
\n" ); document.write( "Next we use a property of logarithms, \"log%28m%2C+%28p%5En%29%29+=+n%2Alog%28m%2C+%28p%29%29\", to \"move\" the exponent out in front:
\n" ); document.write( "\"log%28%28q%29%29+=+x%2Alog%28%28a%29%29\"
\n" ); document.write( "Dividing both sides by log(a):
\n" ); document.write( "\"log%28%28q%29%29%2Flog%28%28a%29%29+=+x\"
\n" ); document.write( "Now we will find an expression for 3/x. Taking the reciprocal of each side:
\n" ); document.write( "\"log%28%28a%29%29%2Flog%28%28q%29%29+=+1%2Fx\"
\n" ); document.write( "Multiplying each side by 3:
\n" ); document.write( "\"3log%28%28a%29%29%2Flog%28%28q%29%29+=+3%2Fx\"

\n" ); document.write( "Now we will use
\n" ); document.write( "\"q+=+b%5Ey\" 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:
\n" ); document.write( "\"q+=+b%5Ey\"
\n" ); document.write( "\"log%28%28q%29%29+=+log%28%28b%5Ey%29%29\"
\n" ); document.write( "\"log%28%28q%29%29+=+y%2Alog%28%28b%29%29\"
\n" ); document.write( "\"log%28%28q%29%29%2Flog%28%28b%29%29+=+y\"
\n" ); document.write( "\"log%28%28b%29%29%2Flog%28%28q%29%29+=+1%2Fy\"
\n" ); document.write( "\"2log%28%28b%29%29%2Flog%28%28q%29%29+=+2%2Fy\"

\n" ); document.write( "Now we will use \"q+=+c%5Ez\" to get an expression for z:
\n" ); document.write( "\"q+=+c%5Ez\"
\n" ); document.write( "\"log%28%28q%29%29+=+log%28%28c%5Ez%29%29\"
\n" ); document.write( "\"log%28%28q%29%29+=+z%2Alog%28%28c%29%29\"
\n" ); document.write( "\"log%28%28q%29%29%2Flog%28%28c%29%29+=+z\"

\n" ); document.write( "Now let's look at what we have for 3/x - 2/y. Substituting in the expressions we found above we should get:
\n" ); document.write( "\"3log%28%28a%29%29%2Flog%28%28q%29%29+-+2log%28%28b%29%29%2Flog%28%28q%29%29\"
\n" ); document.write( "These already have the same denominator so we can subtract:
\n" ); document.write( "\"%283log%28%28a%29%29-+2log%28%28b%29%29%29%2Flog%28%28q%29%29+\"
\n" ); document.write( "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:
\n" ); document.write( "\"%28log%28%28a%5E3%29%29-+log%28%28b%5E2%29%29%29%2Flog%28%28q%29%29+\"
\n" ); document.write( "Still not very interesting. But if we use another property of logarithms, \"log%28m%2C+%28p%29%29+-+log%28m%2C+%28v%29%29+=+log%28m%2C+%28p%2Fv%29%29\", to combine these logs:
\n" ); document.write( "\"log%28%28a%5E3%2Fb%5E2%29%29%2Flog%28%28q%29%29+\"
\n" ); document.write( "Starting to look interesting... From \"a%5E3+=+b%5E2+%2Ac\", we can get \"a%5E3%2Fb%5E2+=+c\". Substituting this in we get:
\n" ); document.write( "\"log%28%28c%29%29%2Flog%28%28q%29%29+\"
\n" ); document.write( "This looks vaguely familiar. Looking back we can see that
\n" ); document.write( "\"log%28%28q%29%29%2Flog%28%28c%29%29+=+z\"
\n" ); document.write( "The left side of this is the reciprocal of what we currently have for 3/x - 2/y! So:
\n" ); document.write( "\"3%2Fx-2%2Fy=1%2Fz\"

\n" ); document.write( "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. \n" ); document.write( "
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