# log change of base...

• Oct 21st 2008, 06:43 PM
namp4namp
log change of base...
• Oct 21st 2008, 06:47 PM
TSizzle55
log (with base (a)) of f(x) is equal to log[f(x)] divided by log(a)
• Oct 21st 2008, 07:36 PM
Soroban
Hello, namp4namp!

Quote:

Solve: .$\displaystyle \log_2(x+1) - \log_4(x) \;=\;1$ .[1]

Let: $\displaystyle \log_4(x) \:=\:P\quad\Rightarrow\quad 4^P \:=\:x$

Take logs (base 2): .$\displaystyle \log_2(4^P) \;=\;\log_2(x) \quad\Rightarrow\quad \log_2(2^2)^P \;=\;\log_2(x)$

. . $\displaystyle \log_2(2^{2P}) \;=\;\log_2(x) \quad\Rightarrow\quad2P\!\cdot\!\underbrace{\log_2 (2)}_{\text{This is 1}} \;=\;\log_2(x) \quad\Rightarrow\quad P \;=\;\tfrac{1}{2}\log_2(x)$
Hence: .$\displaystyle \log_4(x) \;=\;\tfrac{1}{2}\log_2(x)$

Substitute into [1]: .$\displaystyle \log_2(x+1) - \tfrac{1}{2}\log_2(x) \;=\;1$

Multiply by 2: .$\displaystyle 2\log_2(x+1) - \log_2(x) \;=\;2 \quad\Rightarrow\quad 2\log_2(x+1)\;=\;\log_2(x) + \log_2(4)$

. . $\displaystyle \log_2(x+1)^2 \;=\;\log_2(4x) \quad\Rightarrow\quad (x+1)^2 \;=\;4x$

. . $\displaystyle x^2+2x+1 \;=\;4x \quad\Rightarrow\quad x^2 - 2x + 1 \;=\;0$

Factor: .$\displaystyle (x-1)(x-1) \;=\;0 \quad\Rightarrow\quad \boxed{x \:=\:1}$