Find the value of x in the equation 49^x+1 + 72x = 350.

Question

ScanSolve AI Tutor · Accepted Answer

Here's the solution for question 3. a):

Rewrite the equation with a common base.
The base 49 can be written as $7^2$ .
$(7^2)^{x+1} + 7^{2x} = 350$
Apply the exponent rule $(a^m)^n = a^{mn}$ :
$7^{2(x+1)} + 7^{2x} = 350$ $7^{2x+2} + 7^{2x} = 350$
Use the exponent rule $a^{m+n} = a^m \cdot a^n$ .
$7^{2x} \cdot 7^2 + 7^{2x} = 350$
Calculate $7^2$ :
$49 \cdot 7^{2x} + 7^{2x} = 350$
Factor out the common term $7^{2x}$ .
$7^{2x}(49 + 1) = 350$ $7^{2x}(50) = 350$
Isolate the exponential term $7^{2x}$ .
$7^{2x} = \frac{350}{50}$ $7^{2x} = 7$
Equate the exponents.
Since $7 = 7^1$ , we can equate the exponents:
$2x = 1$ $x = \frac{1}{2}$
The value of $x$ is $\boxed{\frac{1}{2}}$ .

Answer