Logarithms

[aleckamadden1990]

I have an exam coming up, and I don't know how to solve a few problems.

1) 4^(logbase2x)(6) = 6 (the answer is x=2)

2) logbasex (logbase1/2)^1/4 = 1 (the answer is x=2

I just don't know how they got the answers

 

[tkhunny]

The second makes no sense. Please give it another go on the presentation.

Logarithms at this level have only a few rules. Do you know them?

4^(logbase2x)(6) = 6

Seems like a logarithm might help.

log(2x)(6) * log(4) = log(6)

Now some arithmetic

log[2x](6) = log(6)/log(4)

Now it looks suspicious. Remember this one? log(a) = log(a)/log(b). Use that on the left hand side and see if anything pops out at you.

 

[aleckamadden1990]

2) log[x](log[1/2](1/4) = 1

 

[tkhunny]

So, the log in the middle has no explicit base? We are to assume Base 10?

How did the first go?

 

[lookagain]

2) log[x](log[1/2](1/4) = 1

It looks to be 1/2 should be the base in the middle and x for the base on the outside:


\(\displaystyle \log_x\bigg[\log_{\frac{1}{2}}\bigg(\dfrac{1}{4}\bigg)\bigg] \ = \ 1\)

 

[soroban]

Hello, aleckamadden1990!

These are tricky ones.
Hope they don't show up on your exam . . .

\(\displaystyle 1)\;\;4^{\log_{2x}(6)} \:=\: 6\)

Take logs, base 4: .\(\displaystyle \log_4\left(4^{\log_{2x}(6)}\right) \:=\:\log_4(6)\)

. . . . . . . . . . . .\(\displaystyle \log_{2x}(6)\cdot\underbrace{\log_4(4)}_{\text{This is 1}} \:=\:\log_4(6)\)

. . . . . . . . . . . . . . . . . . \(\displaystyle \log_{2x}(6) \:=\:\log_4(6)\)


\(\displaystyle \text{Let }\,\log_{2x}(6) =P\;\text{ and }\;\log_4(6) =P\)

\(\displaystyle \text{Then: }\: (2x)^P = 6\;\text{ and }\;4^P = 6\)

\(\displaystyle \text{Hence: }\: (2x)^P \:=\:4^P\)

\(\displaystyle \text{Take the }P^{th}\text{ root: }\:2x \:=\:4\)

\(\displaystyle \text{Therefore: }\: x \:=\:2\)

\(\displaystyle 2)\;\;\log_x\left[\log_{\frac{1}{2}}\left(\frac{1}{4}\right)\right] \:=\:1\)

We have: .\(\displaystyle x^1 \:=\: \log_{\frac{1}{2}}\left(\frac{1}{4}\right)\)

. . . . . \(\displaystyle \left(\frac{1}{2}\right)^x \:=\:\frac{1}{4} \quad\Rightarrow\quad \frac{1}{2^x} \:=\:\frac{1}{4} \quad\Rightarrow\quad 2^x \:=\:4\)

Therefore: .\(\displaystyle x \:=\:2\)