Exponential Function to Undo Log

[Jason76]

1.

\(\displaystyle log_3(x) = 5\)

Find \(\displaystyle x\) by taking both sides to the power of 3.

\(\displaystyle 3^{log_3(x)} = 3^{5}\)

\(\displaystyle x = 243\)

2.

\(\displaystyle log_4(1/4) = y\)

\(\displaystyle 4^{log_4(1/4)} = 4^{y}\)

\(\displaystyle \dfrac{1}{4} = 4^{y}\)

But what if we knew all the variables except for \(\displaystyle b\) in example:\(\displaystyle log_b(243) = 5\)?

How would we find b?

 

[JeffM]

1.

\(\displaystyle log_3(x) = 5\)

Find \(\displaystyle x\) by taking both sides to the power of 3.

\(\displaystyle 3^{log_3(x)} = 3^{5}\) True but ...

\(\displaystyle x = 243\)

2.

\(\displaystyle log_4(1/4) = y\)

\(\displaystyle 4^{log_4(1/4)} = 4^{y}\)

\(\displaystyle \dfrac{1}{4} = 4^{y}\) True but not the answer sought.

But what if we knew all the variables except for \(\displaystyle b\) in example:\(\displaystyle log_b(243) = 5\)?

How would we find b?

You need to rely on the definition of logs to solve these problems. You have learned a formula, but not the idea.

DEFINITION \(\displaystyle Given\ u > 0\ and\ u \ne 1\ and\ v > 0,\ then\ log_u(v) = w \iff v = u^w.\)

So problem 1:

\(\displaystyle log_3(x) = 5 \implies x = 3^5 = 243.\)

And problem 2:

\(\displaystyle y = log_4\left(\dfrac{1}{4}\right) \implies 4^y = \dfrac{1}{4} = \dfrac{1}{4^1} = 4^{(-1)} \implies y = - 1.\)

And your extra problem

\(\displaystyle log_b(243) = 5 \implies b^5 = 243 \implies \sqrt[5]{b^5} = \sqrt[5]{243} \implies b = 3.\)

 

[HallsofIvy]

Do you know that \(\displaystyle \frac{1}{a}= a^{-1}\)

 

[Jason76]

\(\displaystyle log_b(243) = 5 \implies b^5 = 243 \implies \sqrt[5]{b^5} = \sqrt[5]{243} \implies b = 3.\)

Thanks for the help.

 

[lookagain]

1.

\(\displaystyle log_3(x) = 5\)

Find \(\displaystyle x\) by taking both sides to the power of 3. . . . You are not taking both sides to the power of 3. You are having 3 raised to each side (as shown in your next step).
\(\displaystyle 3^{log_3(x)} = 3^{5}\)

\(\displaystyle x = 243\)

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