How do I solve the equation e to the power of x equals 2 using logarithms?

Answer 1

When the logarithm is taken of any number to a power the result is that power times the log of the number; so taking logs of both sides gives:

e^x = 2

→ log(e^x) = log 2

→ x log e = log 2

Dividing both sides by log e gives:

x = (log 2)/(log e)

The value of the logarithm of the base when taken to that base is 1.

The logarithms can be taken to any base you like, however, if the base is e (natural logs, written as ln), then ln e = 1 which gives

x = (ln 2)/1 = ln 2

This is in fact the definition of a logarithm: the logarithm to a specific base of a number is the power of the base which equals that number.

In this case ln 2 is the number x such that e^x = 2.

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This also means that you can calculate logs to any base if you can find logs to a specific base:

log (b^x) = y

→ x log b = log y

→ x = (log y)/(log b)

In other words, the log of a number to a given base, is the log of that number using any [second] base you like divided by the log of the base to the same [second] base.

eg log₂ 8 = ln 8 / ln 2 = 2.7094... / 0.6931... = 3

since log₂ 8 = 3 it means 2³ = 8 (which is true).

Answer 2

If e^x = 2 then taking logarithms to the base e, x = ln(2).