Hence, or otherwise, solve for x if 7^(2x) − 48(7^x) − 49 = 0

Question

ScanSolve AI Tutor · Accepted Answer

Here's how to solve the exponential equation:

1.3.2
To solve $7^{2x} - 48(7^x) - 49 = 0$ , we can use a substitution to transform it into a quadratic equation.

Let $u = 7^x$ .
Then $7^{2x} = (7^x)^2 = u^2$ .
Substitute $u$ into the given equation:
$u^2 - 48u - 49 = 0$
Solve the quadratic equation for $u$ .
We can factor the quadratic expression:
$(u - 49)(u + 1) = 0$
This gives two possible values for $u$ :
$u = 49 \quad \text{or} \quad u = -1$
Substitute back $7^x$ for $u$ and solve for $x$ .
Case 1: $u = 49$
$7^x = 49$
Since $49 = 7^2$ , we have:
$7^x = 7^2$
Therefore, $x = 2$ .

Case 2: $u = -1$
$7^x = -1$
An exponential function with a positive base (like 7) can never produce a negative result. Thus, there is no real solution for $x$ in this case.

The only valid solution for $x$ is from Case 1.

Hence, or otherwise, solve for x if 7^(2x) - 48(7^x) − 49 = 0