Find x for 4^(x-4) = 7.

Answer I got, using log, was (log(7)/2log(2)) + 4

but the actual answer was (ln(7)/2ln(2)) + 4

I plugged both in my calculator and turns out both are the equivalent value.

Anyways, is using either one of ln or log appropriate for this question? Obviously ln is when log has the base e, and log is when it has the base 10.

Final question: How do I know when to use which? that is which of ln or log is used when solving a question??

For example, if a question asks to find x for e^x = 100, I will use ln since ln(e) cancels out.

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

So is using either log or ln the same?

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 $b > 0, b\neq 1$).

September 20, 2019 06:53 AM

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.

September 20, 2019 06:56 AM

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