SOLUTION: Solve the following system for (x,y): 1. log9(x)+ logy(8)= 2 2. logx(9)+ log8(y)= (8/3) The term attached to log is the base, the term in parenthesis the argument. This feel

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve the following system for (x,y): 1. log9(x)+ logy(8)= 2 2. logx(9)+ log8(y)= (8/3) The term attached to log is the base, the term in parenthesis the argument. This feel      Log On


   



Question 566479: Solve the following system for (x,y):
1. log9(x)+ logy(8)= 2
2. logx(9)+ log8(y)= (8/3)
The term attached to log is the base, the term in parenthesis the argument. This feels like it should be simple, but after 30 minutes of useless substituting, I figure I'm missing an obvious connection . . .
Thanks to anyone who sees it.
TS

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I'll connect you.
Let's change variables.
w=log%289%2Cx%29 and z=log%28y%2C8%29
The first equation transforms easily:
log%289%2Cx%29%2Blog%28y%2C8%29=2 --> w%2Bz=2
For the next one, we have to use a property of logarithms.
As with all the properties of logarithms, you can remember it, or you can rediscover it from the definition of logarithm every time you need it.
I always opted for the second choice, but in this case, rediscovery is not that easy, so I say understand the proof once, and then try to remember that
log%28a%2Cb%29=1%2Flog%28b%2Ca%29
To help students remember that property, a popular tutor used to call it "skewer it and turn it upside down." That helped because he would write a very large 1 above the log%28b%2Ca%29 and pretend it was a toothpick skewering an olive, before pretending to turn it upside down and take it to his mouth, and
the log%28a%2Cb%29 he would then write on the board sort of looks like an upside down log%28b%2Ca%29. So
log%28x%2C9%29=1%2Flog%289%2Cx%29 , so log%28x%2C9%29=1%2Fw and
log%288%2Cy%29=1%2Flog%28y%2C8%29 , so log%288%2Cy%29=1%2Fz
With that, the second equation gets transformed
log%28x%2C9%29%2B+log%288%2Cy%29=8%2F3 --> 1%2Fw%2B1%2Fz=8%2F3 --> %28w%2Bz%29%2Fwz=8%2F3
So we have two equations
w%2Bz=2 and %28w%2Bz%29%2Fwz=8%2F3
and just need to find w and z. No logarithm worries (for now).
At this point I would divide the first equation by the second equation
%28w%2Bz%29%2F%28%28%28w%2Bz%29%2Fwz%29%29=2%2F%28%288%2F3%29%29 --> %28w%2Bz%29%2A%28wz%2F%28w%2Bz%29%29=2%2A%283%2F8%29 --> wz=3%2F4%29
If you figure out here that the numbers w and z must be 3%2F2 and 1%2F2, good for you.
I had to remember that they would be solutions of the quadratic equation
x%5E2-2x%2B3%2F4=0 <--> 4x%5E2-8x%2B3=0 and use the quadratic formula.
It could be that w=3%2F2 and z=1%2F2 ,
or that w=1%2F2 and z=3%2F2 .
The first solution:
w=3%2F2 and z=1%2F2
w=3%2F2 --> log%289%2Cx%29=3%2F2 --> x=9%5E%283%2F2%29=%28sqrt%289%29%29%5E3=3%5E3=27
and z=1%2F2 --> log%288%2Cy%29=1%2Fz=1%2F%281%2F2%29=2 --> y=8%5E2=64
The second solution:
w=1%2F2 and z=3%2F2
w=1%2F2 --> w=log%289%2Cx%29=1%2F2--> x=9%5E%281%2F2%29=sqrt%289%29=3
z=3%2F2 --> log%288%2Cy%29=1%2Fz=1%2F%283%2F2%29=2%2F3 --> y=8%5E%282%2F3%29=%28root%283%2C8%29%29%5E2=2%5E2=4