Evaluating Logarithms

[Matty R]

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

1. The problem statement, all variables and given/known data

Evaluate : 2^x - 2^{1-x} = 1


2. Relevant equations

logb^n = nlogb


3. The attempt at a solution

2^x - 2^{1-x} = 1

log(2^x) - log(2^{1-x}) = log(1)

xlog(2) - (1-x)log(2) = 0

log(2)(x - (1-x)) = 0

log(2)(2x-1) = 0

2xlog(2) - 1log(2) = 0

2xlog(2) = log(2)

2x = \frac{log(2)}{log(2)}

2x = 1

x = \frac{1}{2}

[Fightfish]

If you take log on both sides, you get log(2^x - 2^{1-x}) on the LHS, which is not equal to log(2^x) - log(2^{1-x}).
The method to solve it is not by logarithms; try simplifying the expression to see if you can see the underlying equation present.

[Mark44]

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

1. The problem statement, all variables and given/known data

Evaluate : 2^x - 2^{1-x} = 1


2. Relevant equations

logb^n = nlogb


3. The attempt at a solution

2^x - 2^{1-x} = 1

Your next step is wrong. You can take the log of each side of an equation, but there is no property that says log(a - b) = log a - log b. That is what you've done, and it is invalid.

Rewrite your equation as 2x - 2/2x = 1, and then multiply both sides by 2x. That should give you an equation that is quadratic in form (in powers of 2x) that you can solve.


log(2^x) - log(2^{1-x}) = log(1)

xlog(2) - (1-x)log(2) = 0

log(2)(x - (1-x)) = 0

log(2)(2x-1) = 0

2xlog(2) - 1log(2) = 0

2xlog(2) = log(2)

2x = \frac{log(2)}{log(2)}

2x = 1

x = \frac{1}{2}

[Matty R]

Thanks for the replies. :smile:

I'd been doing questions with logarithms just before I got to this question, so when I saw the power of x I thought of logarithms and nothing else.

Sooooo, how about this?

2^x - 2^{1-x} = 1

2^x - 2^1 \cdot 2^{-x} = 1

2^x - \frac{2^1}{2^x} = 1

2^x - \frac{2}{2^x} - 1= 0

y = 2^x

y - \frac{2}{y} - 1 = 0

y^2 - 2 - y = 0

y^2 - y - 2 = 0

y^2 - 2y + 1y - 2 = 0

y(y - 2) + 1(y - 2) = 0

(y - 2)(y + 1) = 0

y = 2 and -1

y = -1

2^x \neq -1 (2^x cannot be negative)

y = 2

2^x = 2

log(2^x) = log(2)

xlog(2) = log(2)

x = \frac{log(2)}{log(2)}

x = 1