# Math Help - Please clarify my workings for a somewhat complex (for me) logarithm question

1. ## Please clarify my workings for a somewhat complex (for me) logarithm question

Hi all,

I've been confronted with a somewhat complex logarithm question in the form of solving for x. After a long time of thinking and trying various methods, I finally came up with two possible methods for solving the problem. However, I am not sure which (or if either) is correct, or if I have made a mistake along the way. Please clarify the correct method.

The question is:

log2(X ^ log2(x)) = 4

From there, I used the power rule and brought log2(x) down the front of the equation:

log2(x) * log2(x) = 4

Then I added the logs together:

log2(x) + log2(x) = 4

log2(2x) = 4

Then I rearranged the equation to indical form:

2 ^ 4 = 2X

16 = 2X

x = 8

Thus, x = 8. However, I want to suggest an alternate method as well that returns to this part of the equation:

log2(x) * log2(x) = 4

Perhaps instead of adding them together like I did in the above working, I simply multiply the two logs together to achieve:

log2(x ^ 2) = 4

Rearrange and solve to equal:

2 ^ 4 = x ^ 2

16 = x ^ 2

x = 4

Any help will be much appreciated!

Thanks,

Nathaniel

2. Originally Posted by BinaryBoy
Hi all,

I've been confronted with a somewhat complex logarithm question in the form of solving for x. After a long time of thinking and trying various methods, I finally came up with two possible methods for solving the problem. However, I am not sure which (or if either) is correct, or if I have made a mistake along the way. Please clarify the correct method.

The question is:

log2(X ^ log2(x)) = 4

[snip]
You require $x^{\log_2(x)} = 16$ (why?).

Therefore $\log_x(16) = \log_2(x)$. By inspection, x = 4.

3. Thanks for the help, Mr Fantastic.

It was actually a practice question for an exam. Never done anything like it before, but I guess its purpose is to extend one's thinking. Which is quite good, of course. So, since x = 4, my second method was correct for working it out?

Thanks again!

Nathaniel

4. Originally Posted by BinaryBoy
Thanks for the help, Mr Fantastic.

It was actually a practice question for an exam. Never done anything like it before, but I guess its purpose is to extend one's thinking. Which is quite good, of course. So, since x = 4, my second method was correct for working it out?

Thanks again!

Nathaniel
$(\log_2(x))^2 \neq \log_2 x^2$ so your working is wrong. However, it got of to a good start - better than mine in fact. Because:

$(\log_2 (x))^2 = 4 \Rightarrow \log_2(x) = \pm 2 \Rightarrow x = 4, \frac{1}{4}$ so there are two solutions. And $x = \frac{1}{4}$ is not necessarily obvious by inspection from my earlier working.

5. Originally Posted by BinaryBoy
Hi all,

I've been confronted with a somewhat complex logarithm question in the form of solving for x. After a long time of thinking and trying various methods, I finally came up with two possible methods for solving the problem. However, I am not sure which (or if either) is correct, or if I have made a mistake along the way. Please clarify the correct method.

The question is:

log2(X ^ log2(x)) = 4

From there, I used the power rule and brought log2(x) down the front of the equation:

log2(x) * log2(x) = 4

Then I added the logs together:

log2(x) + log2(x) = 4
The logs were multiplied, not added! You can't just change a*b to a+ b.

log2(2x) = 4

Then I rearranged the equation to indical form:

2 ^ 4 = 2X

16 = 2X

x = 8

Thus, x = 8. However, I want to suggest an alternate method as well that returns to this part of the equation:

log2(x) * log2(x) = 4

Perhaps instead of adding them together like I did in the above working, I simply multiply the two logs together to achieve:

log2(x ^ 2) = 4

Rearrange and solve to equal:

2 ^ 4 = x ^ 2

16 = x ^ 2

x = 4

Any help will be much appreciated!

Thanks,

Nathaniel