Can someone tell me why $\log_b(c) = s \cdot \log_b(d) \implies c = d$?

Question

I encountered this statement on the U of T calculus preparation website, on the function and inverses section.

Exercise 8 has two hints, the second of which states the following:

If $\log_b(c) = s \cdot \log_b(d)$, then we can conclude that $c=d$ since logarithmic functions are one-to-one

But $\log_{10}(100) = 2 \cdot \log_{10}(10)$, and $100 \neq 10$.

Is there something I'm missing?

Answer 1

It was a typo and it is not true.

On the web site the problem is:

The solution to $2\ln(x)+\ln(x+2)−\ln(x^2+2x)=−ln(2)$ is:

The first hint lists the three basic log identities:

1. $\log_b(r⋅s)=\log_br+\log_bs$
2. $\log_b(\frac rs)=\log_br−\log_bs$
3. $\log_b(r^s)=s\log_br$

Which you are to apply to $2\ln(x)+\ln(x+2)−\ln(x^2+2x)=−ln(2)$ to get

$\ln \frac {x^2(x+2)}{x+2} = \ln x = -\ln(2)$

$\ln x = -\ln 2$

Then the second hint was $ \log_b(c)=s⋅\log_b(d)\implies c = d$. That had a typo. They mean $ \log_b(c)=s\log_b(d)\implies c = d^s$

To which you are to apply $x = 2^{-1} = \frac 12$

Answer 2

Your statement implies $c = d^s$ so the implication would only be true iff $s = 1.$ Also in general, a function $f$ is injective iff $f(x) = f(y)$ implies $x = y.$

Answer 3

$$\log_b(c) = s \cdot \log_b(d)$$ does not imply $$c=d$$ unless $s=1$.

Thus you are correct in detecting the typo in the problem.