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...

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

$\displaystyle log_2(y) = 1$

$\displaystyle y = 2$

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.)

Originally Posted by undefined

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???

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: -

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

$\displaystyle log_2(log_2(log_2(x))) = log_2(2)$ [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.

..

Thanks..

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