Logarithm Problem?? How to find the value of "X"??

May 2010
18
1
Okay here's the problem: -

FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: -

[FONT=&quot][LOG2LOG2LOG2X = 1][/FONT]

The given answer is "X" = 16.

But I have no idea how to find that out. Do I have to write [FONT=&quot]LOG22 instead of 1?

Please help...
(Worried)

Thanks in advance...
[/FONT]
 

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MHF Hall of Honor
Mar 2010
2,340
821
Chicago
Okay here's the problem: -

FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: -

[FONT=&quot][LOG2LOG2LOG2X = 1][/FONT]

The given answer is "X" = 16.

But I have no idea how to find that out. Do I have to write [FONT=&quot]LOG22 instead of 1?

Please help...
(Worried)

Thanks in advance...
[/FONT]

Work from the outside in.

So \(\displaystyle log_2(log_2(log_2(x))) = 1\)

\(\displaystyle log_2(y) = 1\)

\(\displaystyle y = 2\)

Now write

\(\displaystyle log_2(log_2(x)) = 2\)

\(\displaystyle log_2(z) = 2\)

\(\displaystyle z = 4\)

Lastly,

\(\displaystyle log_2(x) = 4\)

\(\displaystyle x = 16\)

Alternatively, you could do \(\displaystyle LHS = RHS \Longrightarrow 2^{LHS} = 2^{RHS}\) three times. (LHS = left hand side, etc.)
 
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May 2010
18
1
Work from the outside in.

So \(\displaystyle log_2(log_2(log_2(x))) = 1\)

\(\displaystyle log_2(y) = 1\)

\(\displaystyle y = 2\)

Now write

\(\displaystyle log_2(log_2(x)) = 2\)

\(\displaystyle log_2(z) = 2\)

\(\displaystyle z = 4\)

Lastly,

\(\displaystyle log_2(x) = 4\)

\(\displaystyle x = 16\)

Alternatively, you could do \(\displaystyle LHS = RHS \Longrightarrow 2^{LHS} = 2^{RHS}\) three times. (LHS = left hand side, etc.)

Thanks...

I've thought of another way, but am not sure if it's valid. Please let me know if it's correct: -

\(\displaystyle log_2(log_2(log_2(x))) = 1\)

\(\displaystyle log_2(log_2(log_2(x))) = log_2(2)\) [I can write \(\displaystyle log_2(2)\) instead of \(\displaystyle 1\), can't I?]

So,

\(\displaystyle log_2(log_2(x)) = 2\) [Now this is what I'm unsure about. Can I eliminate \(\displaystyle log_2\) from both sides in this case??]

Next,

\(\displaystyle log_2(log_2(x)) = 2*log_2(2)\) [where \(\displaystyle log_2(2) = 1\) ]

\(\displaystyle log_2(log_2(x)) = log_2(2)^2 \Longrightarrow log_2(log_2(x)) = log_2(4)\)

\(\displaystyle log_2(x) = 4\)

\(\displaystyle \Longrightarrow log_2(x) = 4*log_2(2)\)

\(\displaystyle \Longrightarrow log_2(x) = log_2(2)^4\)

\(\displaystyle \Longrightarrow log_2(x) = log_2(16)\)

Thus \(\displaystyle x = 16\)

Now is it a valid method???
 

undefined

MHF Hall of Honor
Mar 2010
2,340
821
Chicago
Thanks...

I've thought of another way, but am not sure if it's valid. Please let me know if it's correct: -

\(\displaystyle log_2(log_2(log_2(x))) = 1\)

\(\displaystyle log_2(log_2(log_2(x))) = log_2(2)\) [I can write \(\displaystyle log_2(2)\) instead of \(\displaystyle 1\), can't I?] Yes.

So,

\(\displaystyle log_2(log_2(x)) = 2\) [Now this is what I'm unsure about. Can I eliminate \(\displaystyle log_2\) from both sides in this case??]

Yes you can. This is because the function \(\displaystyle \color{red}f(x)=\log_2(x)\) is one-to-one, or injective, which you probably haven't learned yet. But it means that f(a) = f(b) implies a = b, and graphically, f(x) passes the "horizontal line test."

Next,

\(\displaystyle log_2(log_2(x)) = 2*log_2(2)\) [where \(\displaystyle log_2(2) = 1\) ]

\(\displaystyle log_2(log_2(x)) = log_2(2)^2 \Longrightarrow log_2(log_2(x)) = log_2(4)\) I would write \(\displaystyle \color{red}log_2(2^2)\) so people don't misinterpret as \(\displaystyle \color{red}\left(\log_2(2)\right)^2\).

\(\displaystyle log_2(x) = 4\)

\(\displaystyle \Longrightarrow log_2(x) = 4*log_2(2)\)

\(\displaystyle \Longrightarrow log_2(x) = log_2(2)^4\)

\(\displaystyle \Longrightarrow log_2(x) = log_2(16)\)

Thus \(\displaystyle x = 16\)

Now is it a valid method??? Yes. Good job. :D
..
 
May 2010
18
1
Thanks..

No I haven't learned the " \(\displaystyle f(x) = log_2(x)\) " thing yet.
 
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