Solve: $2^x - 2^{x-2} = 3$

Question

I'm new to this whole forum thing but I am really frustrated.

I am dealing with an exponential function: $$2^x - 2^{x-2} = 3$$

I tried taking the log of each term but been getting {no solution}, since the X's cancel out.

Here's what I have tried: $$2^x - 2^{x-2} = 3\qquad \text{(Take log of each Term)}$$
$$x \log (2) - (x-2) \log (2) = \log (3) \qquad \text{(I divided by $\log (2)$)}$$ Since $\log (2)\not=0$ I divided by $\log (2)$

$$x-(x-2) = (\log (3) / \log (2))$$

And here is where I run into the problem of X's canceling out on the left side of the equation. However, I know the solution is $2$ because If you plug 2 in for $x$, the equation is true. Did I miss a logarithmic rule or is this a special situation? Please explain in detail if it's not too much trouble.

Thanks,

Typing this on my phone...hope that works.

Answer 1

$\log$ is not a linear function so you cannot take log each term.

Instead from the original equation factor out $2^{x-2}$ you get

$$2^{x-2}(2^2-1)=3$$

which implies

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

and therefore

$$x=2$$

Answer 2

Here is what you need to do:

$$2^x-2^{x-2} = 3$$ $$2^x(1-2^{-2}) = 3$$ $$2^x*\frac{3}{4} = 3$$ $$2^x = 4$$

And now you take the logarithm:

$$x=2$$

Answer 3

It it $$2^x-\frac{2^x}{4}=3$$ so $$2^x\cdot \frac{3}{4}=3$$ and we get $$2^x=2^2$$

Answer 4

$log(a-b)$ doesn't equal to $log(a) - log(b)$. Rather than taking log right away, you can do this with exponent first then use logarithm to find $x$.

Hint : $2^{x-2} = \frac{2^{x}}{2^{2}}$

Use logarithm in the end to find $x$ after simplifying things

Answer 5

Let $y = 2^{x-2}$.

Hence, the original equation becomes

$$ 4y - y = 3$$

From which it follows that $y=1$; and thus,

$$x = 2.$$