logarithmic function

[sarey126]

Problem:

1.) log_2_(x+1)-log_4_x=1

The books gives a hint, "change log_4_x to base 2," however I do not know how to change the base! (Answer in back of book is 1). Thank you!

*assume that numbers and quantities between ^...^ are a superscript and those between _...._ are subscripts*

 

[tkhunny]

Generally (with appropriate restrictions on a, b, and c): \(\displaystyle log_{b}(a)\,=\,\frac{log_{c}(a)}{log_{c}(b)}\)
More Specifically: \(\displaystyle log_{4}(x)\,=\,\frac{log_{2}(x)}{log_{2}(4)}\)
Can you simplify?

 

[soroban]

Hello, sarey126!

\(\displaystyle 1)\;\;\log_2(x+1)\,-\,\log_4(x)\:=\:1\)

If you don't know the Base-Change Formula, you can invent your own . . .

Let \(\displaystyle \log_4(x)\,=\,a\;\;\Rightarrow\;\;4^a\,=\,x\)

Take logs (base 2): .\(\displaystyle \log_2(4^a)\:=\:\log_2(x)\) .[1]

. . The left side is: .\(\displaystyle a\cdot\log_2(4)\,=\,a\cdot\log_2(2^2)\,=\,a\cdot(2\cdot\log_2(2))\,=\,a\cdot(2\cdot1)\,=\,2a\)

Equation [1] becomes: .\(\displaystyle 2a\:=\:\log_2(x)\;\;\Rightarrow\;\;a\:=\:\frac{1}{2}\cdot\log_2(x)\)

. . Hence: .\(\displaystyle \log_4(x)\:=\:\frac{1}{2}\cdot\log_2(x)\) .[2]


The original equation is: .\(\displaystyle \log_2(x+1)\,-\,\log_4(x)\;=\;1\)

. . Substitute [2]: .\(\displaystyle \log_2(x+1)\,-\,\frac{1}{2}\cdot\log_2(x)\;=\;1\)

Can you finish it now?