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.

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Jul 23 '19 at 09:21
The error that you made is in assuming $\log{(x+y)} = \log{x}+ \log{y}$ (after Take log of each Term ). A better way to approach this problem is to write $$ 2^x = 2^2 \cdot 2^{x-2} = 4 \cdot 2^{x-2} $$ and to use this term in the solution. — Matti P., Jul 23 '19 at 09:22

score 4 · Answer 1 · answered Jul 23 '19 at 09:24

4

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

answered Jul 23 '19 at 09:24

Dr. Sonnhard Graubner

95,283

score 3 · Answer 2 · answered Jul 23 '19 at 09:23

3

$\log$ 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

$$x=2$$

answered Jul 23 '19 at 09:23

score 3 · Answer 3 · answered Jul 23 '19 at 09:29

$log(a-b)$ 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

score 1 · Answer 4 · answered Jul 23 '19 at 09:24

1

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:

$$x=2$$

answered Jul 23 '19 at 09:24

TheCoolDrop

745

score 0 · Answer 5 · answered Jul 24 '19 at 02:01

0

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

Hence, the original equation becomes

$$ 4y - y = 3$$

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

$$x = 2.$$

answered Jul 24 '19 at 02:01

DDS

3,199
1
8
31

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

5 Answers5