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.
