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
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
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