Is "ln" (natural log) and "log" the same thing if used in this answer?

Solution 1:

You can use any logarithm you want.

As a result of the base change formula $$\log_2(7) = \frac{\log(7)}{\log(2)} = \frac{\ln(7)}{\ln(2)} = \frac{\log_b(7)}{\log_b(2)}$$ so as long as both logs have the same base, their ratio will be the same, regardless of the chosen base (as long as $b > 0, b\neq 1$).

Solution 2:

Either is fine. You can write logarithms in terms of any base that you like with the change of base formula $$\log_ba=\frac{\log_ca}{\log_cb}$$

One thing learned from secondary school is that exponential equations can be solved be rewriting the relationship in logarithmic form. As such, your equation can be rewritten as follows: \begin{align} 4^{x-4}&=7\\ x-4&=\log_47\\ x&=4+\log_47 \end{align}

And since $\log_47$ can be rewritten as $\frac{\log7}{\log4}$ or $\frac{\ln7}{\ln4}$ or $\frac{\log_{999876}7}{\log_{999876}4}$ it does not matter which base of logarithm you use.

Solution 3:

Culturally

• Computer science / programming people tend to use log base $2$
• Mathematicians tend to use log base $e$
• Engineers / physicist / chemists etc. tend to use log base $10$

Writers really should make it explicit the first time they use "$\log$", but they don't always. As others have pointed out, the only difference is a constant factor, and in your case the factors in the numerator and denominator cancelled each other out. So the answer to your question is "If it's in a math context you'll probably see $\ln()$ used."

Heck, even if you were asked to solve $10^{x - 3} = 6$ you'd still see $\ln()$ used, even though it looks like $\log_{10}$ might seem more "natural" for that particular problem. It's just what math people tend to do.

Solution 4:

There's an interesting unstated question here: what counts as an answer?

You can clearly argue that using either $\ln$ or $\log_{10}$ should be acceptable. But in that case $$ x = \log_4(7) + 4 $$ should be just as correct. As @BrianMoehring says in his answer, you can use any logarithm you want.

As for

If a question asks to find $2^x=64$, i will use log since "e" isn't present in the question.

I would just say $x=6$. That's really using $\log_2$, by inspection.