Ytukyg

I'm solving an exponential equation and I would like to check my solution with someone. The result should be $x=1$ which is what I get, but I'm not sure if my math is correct.

$$2^x + 2^{x+1} = -2^{x-1}+7 $$
$$2^x *1 + 2^{x}*2 = -2^{x}:2+7 $$
$$2^x *(1 + 2) = -2^{x}:2+7 $$
$$2^x(3) = -2^{x}:2+7 $$
$$2^x(3) + 2^{x}:2 = +7 $$
Dividing $2$ by $2$ is just multiplying by $frac{1}{2}$
$$2^x(3) + 2^{x}* frac{1}{2} = +7 $$
$$2^x(3 +frac{1}{2}) = +7 $$
$$2^x(frac{7}{2}) = +7^1 $$

Now since I have only two exponents (the x and the 1 belonging to the 7) I can equate them
$$x=1$$

Is this correct?

Edit: I'm just seeing this.

$$2^1(frac{7}{2}) = +7^1 $$
$$2(frac{7}{2}) = +7 $$

Simplify 2 and $frac{7}{2}$

$$7 = 7 $$

I think this is correct then? It is 2am here, sorry if this is very obvious haha

You can only equate the exponents if the bases are equal.

Here is another approach:
begin{align*}
2^x + 2^{x + 1} & = -2^{x - 1} + 7\
2^{x + 1} + 2^x + 2^{x - 1} & = 7\
2^{x - 1}(2^2 + 2 + 1) & = 7\
2^{x - 1}(4 + 2 + 1) & = 7\
2^{x - 1} cdot 7 & = 7\
2^{x - 1} & = 1\
2^{x - 1} & = 2^0
end{align*}
Since the bases are equal, we can equate the exponents. Hence,
begin{align*}
x - 1 & = 0\
x & = 1
end{align*}

Everything you did was correct until you reached the step
$$2^xleft(frac{7}{2}right) = 7$$
If you multiply both sides of this equation by $dfrac{2}{7}$, you obtain
begin{align*}
2^x & = 2\
2^x & = 2^1
end{align*}
Since the bases are the same, we may equate the exponents, which yields
$$x = 1$$
as above.

What you cannot do is equate the exponent of the $2$ and the exponent of the $7$. You obtained the correct answer for the wrong reason. To see why, suppose you had
$$2^x = 8^1$$
By your reasoning, $x = 1$. However, $8 = 2^3$. Hence, what we should obtain is
$$2^x = 2^3$$
Since the bases are the same, we can equate the exponents, which yields $x = 3$, not $x = 1$.