Log - same base, and help

[K.ourt]

I have to get that all into one base, but I am unsure of how to proceed and get them both into base 2, or base 4.



And

log6X - log6(x-1)=1
(the 6 is the base). But I keep cancelling my x's by accident. Eeeh.

 

[soroban]

Hello, K.ourt!

$\displaystyle \log_4(x)\,+\,\log_2(y)\,-\,\log_2(z)$

But I am unsure of how to proceed and get them both into base 2, or base 4.

Let: $\displaystyle \,\log_4(x)\,=\,p$, then: $\displaystyle 4^p\,=\,x$

$\displaystyle \;\;$Take logs (base 2): $\displaystyle \,\log_2(4^p)\,=\,\log_2(x)\;\;\Rightarrow\;\;p\cdot\log_2(4)\,=\,\log_2(x)$

$\displaystyle \;\;$And we have: $\displaystyle p\cdot2\,=\,\log_2(x)\;\;\Rightarrow\;\;p\,=\,\frac{1}{2}\cdot\log_2(x) \,=\,\log_2(x^{\frac{1}{2}})$

$\displaystyle \;\;$That is: $\displaystyle \log_4(x)\,=\,\log_2(\sqrt{x})\;$ **


The problem becomes: \(\displaystyle \,\log_2(\sqrt{x})\,+\,\log_2(y)\,-\,\log_2(z)\;=\;\log_2\L\left(\frac{\sqrt{x}\cdot y}{z}\right)\)

$\displaystyle \log_6(x)\,-\,\log_6(x\,-\,1)\:=\:1$
But I keep cancelling my x's by accident. $\displaystyle \;\;$ That shouldn't happen . . . Watch your algebra!

We have: $\displaystyle \log_6\left(\frac{x}{x\,-\,1}\right)\:=\:1\;\;\Rightarrow\;\;\frac{x}{x\,-\,1}\:=\:6^1$

Then: $\displaystyle \,x\:=\:6(x\,-\,1)\;\;\Rightarrow\;\;x\:=\:6x\,-\,6\;\;\Rightarrow\;\;6\,=\,5x\;\;\Rightarrow\;\;x\,=\,\frac{6}{5}$


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**
Are you familiar with the Base Change Formula? $\displaystyle \;\log_a(x)\:=\:\frac{\log_b(x)}{\log_b(a)}$

Then: $\displaystyle \,\log_4(x) \:= \:\frac{\log_2(x)}{\log_2(4)} \:=\;\frac{\log_2(x)}{2} \:=\:\log_2(x^{\frac{1}{2}})$ . . . etc.