solve, giving your answers as exact fractions, the simultaneous equations:

8^(x) = 4^(2x+3)

log(base2)y = log(base2)x + 4

('^' meaning 'to the power of')

I am stuck, i dont know how to do it! PLease show your method,

Thanks, azza xxxz

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- Oct 2nd 2010, 10:55 AMazzalogs and exponentials question
solve, giving your answers as exact fractions, the simultaneous equations:

8^(x) = 4^(2x+3)

log(base2)y = log(base2)x + 4

('^' meaning 'to the power of')

I am stuck, i dont know how to do it! PLease show your method,

Thanks, azza xxxz - Oct 2nd 2010, 11:00 AMPlato
- Oct 2nd 2010, 11:05 AMskeeter
- Oct 2nd 2010, 11:09 AMazza
sorry i typed it in incorrecty

8^(y) = 4^(2x+3)

log(base2)y = log(base2)x + 4

so that gives you 3y = 4x +6 which i make into y = 4/3x + 2 and I put it into the y of the other equation. but how do i solve it then? - Oct 2nd 2010, 11:13 AMskeeter
y = 16x ... substitute 16x for y into the first equation.

- Oct 2nd 2010, 11:20 AMazza
thankyou very much!

i am getting x=1/8, y=2, but the textbook answer is x=3/22 y=24/11

is the textbook wrong or am i? - Oct 2nd 2010, 11:28 AMskeeter
- Oct 2nd 2010, 11:31 AMazza
ahh finally i understand, thankyou so much!

- Oct 2nd 2010, 12:30 PMazza
I have ONe other question:

log(base 3)(2-3x) = log(base9)(6x^2 - 19x + 2)

Thankyou x - Oct 2nd 2010, 12:42 PMskeeter
- Oct 2nd 2010, 01:35 PMazza
sorry, i dont see where that leads?

- Oct 2nd 2010, 03:00 PMskeeter

finish it - Oct 3rd 2010, 04:06 AMHallsofIvy
Another way of looking at it: [tex]log_a(x)= y if and only if so if then 2- 3x= 3^y[tex] and if then .

But so .

Solve - Oct 3rd 2010, 05:24 AMskeeter