logs and exponentials question

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October 2nd 2010, 09:55 AM
azza

logs 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
October 2nd 2010, 10:00 AM
Plato

Quote:

Originally Posted by azza

solve, giving your answers as exact fractions, the simultaneous equations: 8^(x) = 4^{2x+3}

$8^x=4^{2x+3}$ is the same as $2^{3x}=2^{4x+6}$ .

Equate the exponents.
October 2nd 2010, 10:05 AM
skeeter

Quote:

Originally Posted by azza

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

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

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

$8^x = 4^{2x+3}$

$(2^3)^x = (2^2)^{2x+3}$

$2^{3x} = 2^{4x+6}$

$3x = 4x+6$

I assume the right side of the second equation is $\log_2{x} + 4$ rather than $\log_2(x+4)$ ...

$\log_2{y} = \log_2{x} + 4$

$\log_2{y} = \log_2{x} + \log_2{16}$

$\log_2{y} = \log_2(16x)$

$y = 16x$

can you finish it?
October 2nd 2010, 10:09 AM
azza

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?
October 2nd 2010, 10:13 AM
skeeter

y = 16x ... substitute 16x for y into the first equation.
October 2nd 2010, 10:20 AM
azza

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?
October 2nd 2010, 10:28 AM
skeeter

Quote:

Originally Posted by azza

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?

the textbook is correct ...

$3(16x) = 4x+6$

recheck your algebra.
October 2nd 2010, 10:31 AM
azza

ahh finally i understand, thankyou so much!
October 2nd 2010, 11:30 AM
azza

I have ONe other question:

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

Thankyou x
October 2nd 2010, 11:42 AM
skeeter

Quote:

Originally Posted by azza

I have ONe other question:

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

Thankyou x

note that $\displaystyle \log_9{u} = \frac{\log_3{u}}{\log_3{9}}$

next time start a new problem w/ a new thread.
October 2nd 2010, 12:35 PM
azza

sorry, i dont see where that leads?
October 2nd 2010, 02:00 PM
skeeter

$\log_3(2-3x) = \log_9(6x^2 - 19x + 2)$

$\displaystyle \log_3(2-3x) = \frac{\log_3(6x^2 - 19x + 2)}{\log_3{9}}$

$\displaystyle \log_3(2-3x) = \frac{\log_3(6x^2 - 19x + 2)}{2}<br />$

$\log_3(2-3x) = \log_3 {\sqrt{6x^2 - 19x + 2}}$

finish it
October 3rd 2010, 03:06 AM
HallsofIvy

Another way of looking at it: [tex]log_a(x)= y if and only if $x= a^y$ so if $y= log_3(2- 3x)$ then 2- 3x= 3^y[tex] and if $y= log_9(6x^2- 19x+ 2)$ then $6x^2- 19x+ 2= 9^y$ .

But $9= 3^2$ so $6x^2- 19x+ 2= (3^2)^y= 3^{2y}= (3^y)^2= 2- 3x$ .

Solve $6x^2- 19x+ 2= 2- 3x$
October 3rd 2010, 04:24 AM
skeeter

Quote:

Originally Posted by HallsofIvy

Another way of looking at it: [tex]log_a(x)= y if and only if $x= a^y$ so if $y= log_3(2- 3x)$ then 2- 3x= 3^y[tex] and if $y= log_9(6x^2- 19x+ 2)$ then $6x^2- 19x+ 2= 9^y$ .

But $9= 3^2$ so $6x^2- 19x+ 2= (3^2)^y= 3^{2y}= (3^y)^2= 2- 3x$ .

Solve $6x^2- 19x+ 2= 2- 3x$

the last equation should be ...

$6x^2- 19x+ 2= (2- 3x)^2$