Logarithmic help

[Adverb]

Yeah, homework problem that I by no means can remember how to do. Answer is in the back of the book, but I need to know how to get the answer on my own and without a calculator.

LOG(X²-3)=2LOG(X-1)



If I'm not mistaken, it's the same as LOG(X²-3)=LOG((X-1)²), but even if I'm right with that, I can't recall how I can go from there to figure out what X is equal to.

 

[Loren]

LOG(X²-3)=LOG((X-1)²) is correct.

There is a much easier approach than what I am going to describe. Maybe you can discover what that is after following this reasoning.

Let LOG(X²-3) = m. By the definition of log, 10[sup:2z32oi8t]m[/sup:2z32oi8t] = X²-3.
Let LOG((X-1)²) = n. Then 10[sup:2z32oi8t]n[/sup:2z32oi8t] = (X-1)².

But m=n. Therefore 10[sup:2z32oi8t]m[/sup:2z32oi8t] = 10[sup:2z32oi8t]n[/sup:2z32oi8t].

Does this imply that X²-3 = (X-1)² ?

 

[soroban]

Hello, Adverb!

\(\displaystyle \text{Solve for }x\!:\quad\log(x^2-3) \:=\:2\log(x-1)\)

\(\displaystyle \text{If I'm not mistaken, it's the same as: }\:\log(x^2-3)\:=\:\log(x-1)^2\)

You're absolutely correct!


Here's a baby-talk approach to a certain log situation.

\(\displaystyle \text{If we have: }\:\log_b(x) \:=\:\log_b(y)\quad\hdots\;\text{two equal logs (same base)}\)

. . \(\displaystyle \text{we can "un-log" both sides: }\:x \,=\,y\)


\(\displaystyle \text{So you have: }\:\log(x^2-3) \:=\:\log(x-1)^2\)

\(\displaystyle \text{Un-log: }\:x^2-3 \:=\x-1)^2 \quad\Rightarrow x^2-3 \:=\:x^2-2x+1\)

\(\displaystyle \text{Therefore: }\:2x\,=\,4 \quad\Rightarrow\quad\boxed{ x \,=\,2}\)

 

[Adverb]

Thanks you guys, especially Loren for responding so quickly.

 

[lookagain]

r.

LOG(X²-3)=2LOG(X-1)

If I'm not mistaken, it's the same as LOG(X²-3)=LOG((X-1)²),

Adverb, they are not equivalent equations. In the first equation \(\displaystyle (X - 1)\) in \(\displaystyle 2LOG(X - 1)\)

must be positive. But in the second equation, \(\displaystyle (X - 1)\) in \(\displaystyle LOG(X - 1)^2\)may be negative

or positive, because you're squaring it before taking the logarithm.

 

[Deleted member 4993]

Lookagain is correct - however, you can do the step you have indicated. It can introduce extraneous solution just like in:

\(\displaystyle \sqrt{x+1} \ = \ x\)

You'll just have to check the solution to make sure that those are within the domain of original functions.

 

[mmm4444bot]

log(x - 1)^2 = log(x - 1) * log(x - 1)

I'm using function notation and the order of operations, here.

So far, this thread shows that Adverb and Loren do, too, and that Soroban and lookagain don't.

 

[BigGlenntheHeavy]

\(\displaystyle log_{10}(x^2-3) \ = \ 2log_{10}(x-1).\)

\(\displaystyle log_{10}(x^2-3) \ = \ log_{10}(x-1)^2.\)

\(\displaystyle (x^2-3) \ = \ 10^{log_{10}(x-1)^2}.\)

\(\displaystyle x^2-3 \ = \ (x-1)^2 \ = \ x^2-2x+1.\)

\(\displaystyle -2x \ = \ -4, \ \implies \ x \ = \ 2.\)

\(\displaystyle I'll \ leave \ the \ check \ up \ to \ whomever.\)