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

**arijit2005** · May 27th 2010, 06:36 AM

Okay here's the problem: -

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

[LOG2LOG2LOG2X = 1]

The given answer is "X" = 16.

But I have no idea how to find that out. Do I have to write LOG22 instead of 1?

Please help...

Thanks in advance...

**undefined** · May 27th 2010, 07:42 AM

Originally Posted by arijit2005

Okay here's the problem: -

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

[LOG2LOG2LOG2X = 1]

The given answer is "X" = 16.

But I have no idea how to find that out. Do I have to write LOG22 instead of 1?

Please help...

Thanks in advance...

Work from the outside in.

So $log_2(log_2(log_2(x))) = 1$

$log_2(y) = 1$

$y = 2$

Now write

$log_2(log_2(x)) = 2$

$log_2(z) = 2$

$z = 4$

Lastly,

$log_2(x) = 4$

$x = 16$

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

**arijit2005** · May 27th 2010, 08:11 AM

Originally Posted by undefined

Work from the outside in.

So $log_2(log_2(log_2(x))) = 1$

$log_2(y) = 1$

$y = 2$

Now write

$log_2(log_2(x)) = 2$

$log_2(z) = 2$

$z = 4$

Lastly,

$log_2(x) = 4$

$x = 16$

Alternatively, you could do $LHS = RHS \Longrightarrow 2^{LHS} = 2^{RHS}$ 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: -

$log_2(log_2(log_2(x))) = 1$

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

So,

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

Next,

$log_2(log_2(x)) = 2*log_2(2)$ [where $log_2(2) = 1$ ]

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

$log_2(x) = 4$

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

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

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

Thus $x = 16$

Now is it a valid method???

**undefined** · May 27th 2010, 08:26 AM

Originally Posted by arijit2005

Thanks...

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

$log_2(log_2(log_2(x))) = 1$

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

So,

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

Yes you can. This is because the function $\color{red}f(x)=\log_2(x)$ 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,

$log_2(log_2(x)) = 2*log_2(2)$ [where $log_2(2) = 1$ ]

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

$log_2(x) = 4$

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

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

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

Thus $x = 16$

Now is it a valid method??? Yes. Good job.

..

**arijit2005** · May 27th 2010, 08:33 AM

Thanks..

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

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