# exponetial eqaution

• Oct 29th 2009, 12:19 PM
scrible
exponetial eqaution
Hi, I have some equation that I am having trouble solving. (1) $2(2^{2x})-5(2^x)+2=0$ must find x

(2) $4^x-6(2^x)-16=0$
(3) $3^{2x+1} -26(3^x)-9=0$ Must also find x for both of them. and

I also tried this one and came up with the answer y=1 (4) $\log_{2}x+\log_{x}2=2$ I have to ind x
• Oct 29th 2009, 12:41 PM
e^(i*pi)
Quote:

Originally Posted by scrible
Hi, I have some equation that I am having trouble solving. (1) $2(2^{2x})-5(2^x)+2=0$ must find x

(2) $4^x-6(2^x)-16=0$
(3) $3^{2x+1} -26(3^x)-9=0$ Must also find x for both of them. and

I also tried this one and came up with the answer y=1 (4) $\log_{2}x+\log_{x}2=2$ I have to ind x

1. Note that this is a quadratic in $2^x$. $2(2^x)^2-5(2^x)+2=0$ Use the quadratic formula to find 2^x and hence x. As $2^x > 0$ discard any negative solution to the quadratic.

2. $4^x = (2^2)^x = 2^{2x}$. This can then be solved in the same way as above

3. $3^{2x+1} = 3^{2x}\cdot 3^1 = 3\cdot 3^{2x}$ which can also be solved in the same way as above

4. I'm confused, what are you trying to ask?

$log_2(x)+log_x(2)=2$

Use the change of base rule to give you the same base: $log_b(a) = \frac{log_c(a)}{log_c(b)}$. I use base e below but any base (that isn't 1 or 0) will work

$log_2(x) = \frac{ln(x)}{ln(2)}$

$log_x(2) = \frac{ln(2)}{ln(x)}$

$\frac{ln(x)}{ln(2)} + \frac{ln(2)}{ln(x)} = 2$

You can then treat that as a normal fraction question
• Oct 29th 2009, 12:51 PM
masters
Quote:

Originally Posted by scrible
Hi, I have some equation that I am having trouble solving. (1) $2(2^{2x})-5(2^x)+2=0$ must find x

(2) $4^x-6(2^x)-16=0$
(3) $3^{2x+1} -26(3^x)-9=0$ Must also find x for both of them. and

I also tried this one and came up with the answer y=1 (4) $\log_{2}x+\log_{x}2=2$ I have to ind x

Hi scrible,

Here's a start. Maybe you can use this as a model for the next two.

$2(2^{2x})-5(2^x)+2=0$

$(2(2^x)-1)(2^x-2)=0$

$2(2^x)-1=0 \ \ or \ \ 2^x-2=0$

$2(2^x)=1 \ \ or \ \ 2^x=2$

$2^x=\frac{1}{2} \ \ or \ \ 2^x=2$

$2^x=2^{-1} \ \ or \ \ 2^x=2^1$

$x=-1 \ \ or \ \ x=1$
• Oct 29th 2009, 02:24 PM
scrible
Quote:

Originally Posted by masters
Hi scrible,

Here's a start. Maybe you can use this as a model for the next two.

$2(2^{2x})-5(2^x)+2=0$

$(2(2^x)-1)(2^x-2)=0$

$2(2^x)-1=0 \ \ or \ \ 2^x-2=0$

$2(2^x)=1 \ \ or \ \ 2^x=2$

$2^x=\frac{1}{2} \ \ or \ \ 2^x=2$

$2^x=2^{-1} \ \ or \ \ 2^x=2^1$

$x=-1 \ \ or \ \ x=1$

Hi it is me again I worked out this problem $3^{2x+1}-26(3^x)-9=0$
This is actual working:

$3(3^{2x}-26(3^x)-9=0$

then I said let $3^x= y$

thus $3y^2-26y-9=0$

$(3y+1)(y-9)$

$y= -{\frac {1}{3}}$ and $y= 9$
Did I do every thing correctly, and is there anything more that I need ot do? The method seem correct but I am still not getting the answer...(Worried)
• Oct 29th 2009, 03:17 PM
e^(i*pi)
Quote:

Originally Posted by scrible
Hi it is me again I worked out this problem $3^{2x+1}-26(3^x)-9=0$
This is actual working:

$3(3^{2x}-26(3^x)-9=0$

then I said let $3^x= y$

thus $3y^2-26y-9=0$

$(3y+1)(y-9)$

$y= -{\frac {1}{3}}$ and $y= 9$
Did I do every thing correctly, and is there anything more that I need ot do? The method seem correct but I am still not getting the answer...(Worried)

You're correct at the moment. What you need to do next is to recall your original substitution. Also be aware that because of the nature of exponents $3^x > 0$ (0 is a horizontal asymptote) which means one of your solutions is not valid