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

#### arijit2005

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?

(Worried)

[/FONT]

#### undefined

MHF Hall of Honor
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?

(Worried)

[/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.)

• arijit2005

#### arijit2005

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
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. ..

#### arijit2005

Thanks..

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

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