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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