Examination questions

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October 20th 2009, 12:58 PM
deadmotor

Examination questions

i can't solve this question

if 5^x x 25^2y=1 and 3^5x x 9^y = 1/9 calculate the value of x and y
October 20th 2009, 01:10 PM
pickslides

Can you confrim this to be the system?

$5^{x} \times 25^{2y}=1$

$3^{5x} \times 9^{y} = \frac{1}{9}$
October 20th 2009, 01:12 PM
deadmotor

yeah
how did you do that?
October 20th 2009, 01:51 PM
pickslides

I made the equations look all pretty using LaTeX. For further information look at

http://www.mathhelpforum.com/math-he...-tutorial.html

I'll kick this off for you.

$5^{x} \times 25^{2y}=1$ ....(1)

$3^{5x} \times 9^{y} = \frac{1}{9}$ ....(2)

Let's have a look at (1)

$5^{x} =25^{-2y}$

$5^{x} =(5^2)^{-2y}$

$5^{x} =(5^2)^{-2y}$

$5^{x} =5^{-4y}$

$x= -4y$

Aim to do the same thing with (2) and you should find a solution.
October 21st 2009, 02:42 AM
HallsofIvy

here's another way to do that (actually harder- pickslide's method is more fundamental and better):

Given $5^x25^y= 1$ and $3^{5x}9^y= 1$ , take logarithms of both sides (to any base). log(5^x)+ log(25^y)= log(x) or xlog(25)+ ylog(25)= 0 and since [tex]log(25)= log(5^2)= 2log(5)[tex], x log(5)+ 2y log(5)= 0 and, dividing both sides by log(25), x+ 2y= 0.

Similarly, taking logarithms of both sides of $3^{5x}9^y= 1$ , 5x log(3)+ y log(9)= 5x log(3)+ 2y log(3)= 0 so 5x+ 2y= 0. And it should be pretty obvious that x= 0, y= 0 is the only solution to that.

To see the "Latex" code for $5^x25^y= 1$ and $3^{5x}9^y= 1$ , click on the formulas.