(Insert pun about the word "log")

Algebra Level 5

If $f(nx) = \log_n x,$ for some positive real number $n \neq 1,$ then $f(x) =$ ?

This problem is not original.

\log_n x - 1

log_x n

Infinitely many functions

\log_n x

Undefined

log_n x + 1

Finitely many functions

\log x

2 solutions

Steven Yuan
Jan 26, 2015

Let $a = nx,$ so $x = \frac{a}{n}.$

$f(a) = f(nx) = \log_n x = \log_n \frac{a}{n} = \log_n a - 1.$

Thus, $f(x) = \boxed{\log_n x - 1}.$

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I do not agree, you also can assume that f(x) is equal to f(1*x) what means that according to the enunciate is non existing. So a value of n must be included as a premise. Smart attempt anyway.

Mariano PerezdelaCruz - 6 years, 4 months ago

Well, it is stated in the problem that $n \neq 1.$

Steven Yuan - 6 years, 4 months ago

Sorry but the function its not defined nor for n=0 neither for n<0 which are real number too. And what about for x negative? it is real not?.

Mariano PerezdelaCruz - 6 years, 4 months ago

@Mariano PerezdelaCruz – Positive real number means $n > 0$ .

Steven Yuan - 6 years, 4 months ago

@Steven Yuan – But if you stated that the adjective positive is applied to n, means that x it could be negative then the enunciate is ill posted in my opinion. The first question in the enunciate is n the coefficient of x? if so f(x) could be f(1x) or could be f(ax/a ) or f(2ax/2a) that can be interpreted as log{2}x-1, log{2a}x-1,or log{a}x-1. The enunciate should put a comma between the n and x f(n,x)

Mariano PerezdelaCruz - 6 years, 4 months ago

@Mariano PerezdelaCruz – Your first issue is correct; $x$ should be positive.

For your second issue, $n$ is the coefficient of $x$ . All that is given is that $f(nx) = \log_n x$ , for a specific value of $n$ . We don't know that $f(mx) = \log_m x$ for $m \neq n$ . So, we can't say that $f(x) = f \left (a \times \frac{x}{a} \right ) = \log_{a} x - 1,$ unless $n = a$ (which is actually the solution to the problem!).

Steven Yuan - 6 years, 4 months ago

Might not, y = |log_{n} x^{-1}| be the function???

Ninad Jadkar - 6 years, 4 months ago

Can you explain?

Steven Yuan - 6 years, 4 months ago

Poor logic. Here's why..

You said that: $a=nx$ ---(1)

and

$f(a)=\log_n {a} - 1$ ---(2)

Then, you put $a=x$ and claimed that $f(x)=\log_n {x} - 1$ (from $eq^n 2$ )

But look at $eq^n 1$ ...

Putting $a=x$ gives $n=1$ , which is not allowed.

So the function isn't defined.

Somesh Singh - 4 years, 5 months ago

Lu Chee Ket
Jan 29, 2015

y = n x

x = y/ n

f (y) = Log (y/ n)

f(x) = Log (x/ n) = Log x - Log n = Log x - 1 {For Base n.}

Check: f (n x) = Log n x - 1 = Log x {Which is correct.}

f(x) = Log x/ Log n - 1

In the fourth step, you put y=x,

which is wrong, because it means that n=1 from the first step; which is not allowed!

Somesh Singh - 4 years, 5 months ago

(Insert pun about the word "log")

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