- #1
Matty R
- 83
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Hello
I was wondering if someone could help me.
I've got a question on evaluating logarithms, but I've not done anything like this before. I'm only used to the basic logarithmic stuff, and I still find that a bit confusing.
So, I've been trying to do this question for hours now, and keep getting x=1/2, x = -2 or 1 = 1, despite knowing that x = 1
Could anyone shed some light on this for me? I'm so confused.
Thank you
[tex]Evaluate : 2^x - 2^{1-x} = 1[/tex]
[tex]logb^n = nlogb[/tex]
[tex]2^x - 2^{1-x} = 1[/tex]
[tex]log(2^x) - log(2^{1-x}) = log(1)[/tex]
[tex]xlog(2) - (1-x)log(2) = 0[/tex]
[tex]log(2)(x - (1-x)) = 0[/tex]
[tex]log(2)(2x-1) = 0[/tex]
[tex]2xlog(2) - 1log(2) = 0[/tex]
[tex]2xlog(2) = log(2)[/tex]
[tex]2x = \frac{log(2)}{log(2)}[/tex]
[tex]2x = 1[/tex]
[tex]x = \frac{1}{2}[/tex]
I was wondering if someone could help me.
I've got a question on evaluating logarithms, but I've not done anything like this before. I'm only used to the basic logarithmic stuff, and I still find that a bit confusing.
So, I've been trying to do this question for hours now, and keep getting x=1/2, x = -2 or 1 = 1, despite knowing that x = 1
Could anyone shed some light on this for me? I'm so confused.
Thank you
Homework Statement
[tex]Evaluate : 2^x - 2^{1-x} = 1[/tex]
Homework Equations
[tex]logb^n = nlogb[/tex]
The Attempt at a Solution
[tex]2^x - 2^{1-x} = 1[/tex]
[tex]log(2^x) - log(2^{1-x}) = log(1)[/tex]
[tex]xlog(2) - (1-x)log(2) = 0[/tex]
[tex]log(2)(x - (1-x)) = 0[/tex]
[tex]log(2)(2x-1) = 0[/tex]
[tex]2xlog(2) - 1log(2) = 0[/tex]
[tex]2xlog(2) = log(2)[/tex]
[tex]2x = \frac{log(2)}{log(2)}[/tex]
[tex]2x = 1[/tex]
[tex]x = \frac{1}{2}[/tex]