What is the value of log(81) 3?

Question

mathmale · Answer

You'll need several pieces of background info to understand this question.

A number such as 81 is a perfect square, and also a perfect cube.  Can you re-write 81 first as a perfect square and then as a perfect cube?

anonymous · Answer

I have no idea. All i know is log(a)b=c--->a^c=b so
81(?)=3

anonymous · Answer

a. 3
b.¼
c. 4
d. 1/3 
@mathmale

mathmale · Answer

@tae993 :  When you're tempted to respond with, "I have no idea," please take the question as a challenge and look up the topic in question on the Internet (or in your textbook).

You could, for example, quickly make up a table of squares, a table of cubes and a table of fourth powers:


  x        x^2
---    -----

  0          0
  1          1
   2        4
 
and so on, and

x      x^3
--    ---
0        0
1        1
2        8

and so on; please add three more rows to this last table.

You do know something truly relevant and important here:  "All i know is log(a)b=c--->a^c=b"

$$\log _{81}=3$$, what is your base, a?

You'll still need to be able to answer my questions:  What is the square root of 81 and what is the fourth root of 81?


 All i know is log(a)b=c--->a^c=b so
81(?)=3

anonymous · Answer

I have looked on the internet and in my book. I don't understand it. I'm more of a hands-on learner. So its hard for me to understand when i just look at a book. The square root of 81 is 9 the fourth square root is 3.
My base is 81. 
x        x^4
__      ______
0          0
1          1
2          12
x        x^5
__      ______
0          0
1          1
2          16    ? or does the x change from 6 to 8...

anonymous · Answer

@mathmale

mathmale · Answer

Where you have " the fourth square root is 3," please write it as "the fourth root of 81 is 3"...No need to include the words "square root."

But right you are.  The fourth root of 81 is indeed 3.

Therefore , your expression

$$ \log(81) 3$$

mathmale · Answer

becomes y=(log to the base 81) of 3, or $$y=\log _{81}3$$

mathmale · Answer

and you may now substitute 3^4 for 81.  Would you please rewrite your expression with 3^4 substituted for 81?  Use the Draw utility, below, for greater clarity, if you wish.

anonymous · Answer

Would it be 1/4 because 3 is 1/4 of 8? @mathmale

anonymous · Answer

81=3^4 
81^(1/4)=3

mathmale · Answer

thank you for being willing to share your thinking (as you have), even if you're not sure you're right.

"Would it be 1/4 because 3 is 1/4 of 8?"  needs re-writing:  "Wouldn't 81 be 3^4?"  or "Wouldn't 81 raised to the power (1/4) be 3?"

So, your$$y=\log_{81}3 $$
becomes, after substitution, 
$$y=\log_{3^4}3 $$

mathmale · Answer

Think about this.  Can you now see / find a way to find the value of y?
Use your "All i know is log(a)b=c--->a^c=b."

mathmale · Answer

You're on the right track!!

mathmale · Answer

Hint:  rewrite $$y=\log_{3^4}3^1 $$ as an exponential expression with 3^4 as its base.

anonymous · Answer

I think I'm finally getting the hang of this math stuff haha :D its hard learning online for me. 
Sooooo. 1/4=log(81)3?

anonymous · Answer

oh 81=log3/4^3? I'm confused

mathmale · Answer

Your "1/4=log(81)3" is great.  I'd prefer you write this as y = 1/4.

Let's check our work:

$$y=\log_{81}3=\log_{3^4}3^1  $$

mathmale · Answer

Then y = $$y=\log_{3^4}3^1$$

mathmale · Answer

So...$$(3^4)^y=3^1$$

mathmale · Answer

Can you solve this for y?

anonymous · Answer

3*3*3*3=81 so (3^4)^81?
y=81?

anonymous · Answer

@mathmale