# Math Question

**URL:** <https://boards.straightdope.com/t/math-question/731229>\
**Category:** Factual Questions\
**Created:** [September 14, 2015, 6:26pm UTC](https://boards.straightdope.com/t/math-question/731229 "2015-09-14T18:26:15Z")\
**Posts on this page:** 4\
**Page:** 1

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**Author:** ![Rain\_Soaked](https://avatars.discourse-cdn.com/v4/letter/r/e19adc/32.png) [@Rain\_Soaked](https://boards.straightdope.com/u/Rain_Soaked)\
**Post date:** [September 14, 2015, 6:26pm UTC](https://boards.straightdope.com/t/math-question/731229/1 "2015-09-14T18:26:15Z")

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My son gave me this question to solve for **r** :

3[sup]3log[sub]9[/sub]4[/sup] = r[sup]1/2[/sup]

It was given to him by a friend in university.

Solving for r is straightforward. I have tried to calculate the actual value for r by pen and paper without success. Using a calculator gives the answer, r = 64.

Is there a way to simplify this down without a calculator?

And, why yes, I am quite proud of my subscript/superscript coding.

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**Author:** ![leahcim](https://avatars.discourse-cdn.com/v4/letter/l/b4bc9f/32.png) [@leahcim](https://boards.straightdope.com/u/leahcim)\
**Post date:** [September 14, 2015, 6:47pm UTC](https://boards.straightdope.com/t/math-question/731229/2 "2015-09-14T18:47:19Z")

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> [@Rain\_Soaked](#):
>
> My son gave me this question to solve for **r** :
> 
> 3[sup]3log[sub]9[/sub]4[/sup] = r[sup]1/2[/sup]

3[sup]3log[sub]9[/sub]4[/sup]  
= 3[sup]log[sub]9[/sub]64[/sup]  
= 3[sup]log[sub]3[/sub]64/log[sub]3[/sub]9[/sup]  
= 3[sup]log[sub]3[/sub]64/log[sub]3[/sub]9[/sup]  
= 3[sup]log[sub]3[/sub]64/2[/sup]  
= 3[sup]log[sub]3[/sub]64[sup]1/2[/sup][/sup]  
= 64[sup]1/2[/sup]  
= r[sup]1/2[/sup] iff r = 64

Edited to add: I guess the important step is the [change of base formula](http://www.purplemath.com/modules/logrules5.htm) on the third line.

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**Author:** ![Rain\_Soaked](https://avatars.discourse-cdn.com/v4/letter/r/e19adc/32.png) [@Rain\_Soaked](https://boards.straightdope.com/u/Rain_Soaked)\
**Post date:** [September 14, 2015, 6:54pm UTC](https://boards.straightdope.com/t/math-question/731229/3 "2015-09-14T18:54:10Z")

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Thank you!

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**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\
**Post date:** [September 14, 2015, 6:59pm UTC](https://boards.straightdope.com/t/math-question/731229/4 "2015-09-14T18:59:31Z")

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> [@Rain\_Soaked](#):
>
> My son gave me this question to solve for **r** :
> 
> 3[sup]3log[sub]9[/sub]4[/sup] = r[sup]1/2[/sup]

**leahcim** gives the correct answer, and what I say will only be a re-framing of this:

Since the right-hand side is the square root of what we are seeking, we may as well begin by squaring both sides. We can conceptualize this as each factor of 3 on the left-hand side becoming a factor of 3[sup]2[/sup] (i.e., 9) instead. Thus, we get:

9[sup]3log[sub]9[/sub]4[/sup] = r.

At this point, you likely already know how to deal with the left-hand side: 9[sup]log[sub]9[/sub]4[/sup] is, by definition, 4, and since we actually have 3 times this many factors of 9 on our left-hand side, we end up with the product of 3 many 4s:

4[sup]3[/sup] = r.