Evaluating Logarithms: Solving 2^x - 2^{1-x} = 1 in Simple Steps

[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

Homework Statement

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

Homework Equations

$$logb^n = nlogb$$

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

Homework Statement

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

Homework Equations

$$logb^n = nlogb$$

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

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)

$$y = 2$$

$$2^x = 2$$

$$log(2^x) = log(2)$$

$$xlog(2) = log(2)$$

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

$$x = 1$$