# logarithm problem

• Dec 19th 2007, 11:20 PM
afeasfaerw23231233
logarithm problem
logarithm problem
• Dec 20th 2007, 12:15 AM
CaptainBlack
Quote:

Originally Posted by afeasfaerw23231233
logarithm problem

The first question can be transformed into a quadratic in $2^x+2^{-x}$, by observing that:

$(2^x+2^{-x})^2=2^{2x}+2+2^{-2x}=[4^x+4^{-x}]+2$

So the equation becomes:

$
2[2^x+2^{-x}]^2 -7[2^x+2^{-x}]=8=0
$

Now solve this quadratic in $2^x+2^{-x}$ and proceed from there.

RonL
• Dec 20th 2007, 12:18 AM
earboth
Quote:

Originally Posted by afeasfaerw23231233
logarithm problem

Hello,

$(2^x+2^{-x})^2=2^{2x}+2+2^{-2x} = (4^x+4^{-x})+2$

Therefore use $(2^x+2^{-x}) = y$. Then $(4^x+4^{-x}) = y^2-2$ . Your equation becomes:

$2(y^2-2)-7y+10=0$ . Now continue. Don't forget to re-substitute to calculate x.

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

With the second problem you can use the property of logarithms:

$\log_b(a)=c~\implies~\log_a(b)=\frac1c~\implies~\f rac1{\log_a(b)} = c$
• Dec 20th 2007, 05:54 AM
Soroban
Hello, afeasfaerw23231233!

Quote:

$\log_x4 - \log_2x \:=\:1$
From the Base-Change Formula: . $\log_x4 \:=\:\frac{\log_24}{\log_2x}\:=\:\frac{2}{\log_2x}$

The equation becomes: . $\frac{2}{\log_2x} - \log_2x \:=\:1$

. . which simplifies to: . $\left(\log_2x\right)^2 + \log_2x - 2 \:=\:0$

. . which factors: . $\left(\log_2x + 2\right)\left(\log_2x - 1\right) \:=\:0$

Then: . $\log_2x + 2 \:=\:0\quad\Rightarrow\quad \log_2x \:=\:-2$ . . . no real roots

.And: . $\log_2x - 1\:=\:0\quad\Rightarrow\quad \log_2x \:=\:1\quad\Rightarrow\quad \boxed{x \:=\:2}$