Solve for x 5^(2x)+42=13(5^x)

$5^{2 x} + 42 = 13 (5^{x})$

Step 1

Subtract $13 \cdot 5^{x}$ from both sides of the equation.

$5^{2 x} + 42 - 13 \cdot 5^{x} = 0$

Step 2

Rewrite $5^{2 x}$ as exponentiation.

${(5^{x})}^{2} + 42 - 13 \cdot 5^{x} = 0$

Step 3

Substitute $u$ for $5^{x}$ .

$u^{2} + 42 - 13 u = 0$

Step 4

Move $42$ .

$u^{2} - 13 u + 42 = 0$

Step 5

Solve for $u$ .

Factor $u^{2} - 13 u + 42$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $42$ and whose sum is $- 13$ .

$- 7, - 6$

Write the factored form using these integers.

$(u - 7) (u - 6) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$u - 7 = 0$

$u - 6 = 0$

Set $u - 7$ equal to $0$ and solve for $u$ .

Set $u - 7$ equal to $0$ .

$u - 7 = 0$

Add $7$ to both sides of the equation.

$u = 7$

Set $u - 6$ equal to $0$ and solve for $u$ .

Set $u - 6$ equal to $0$ .

$u - 6 = 0$

Add $6$ to both sides of the equation.

$u = 6$

The final solution is all the values that make $(u - 7) (u - 6) = 0$ true.

$u = 7, 6$

Step 6

Substitute $7$ for $u$ in $u = 5^{x}$ .

$7 = 5^{x}$

Step 7

Solve $7 = 5^{x}$ .

Rewrite the equation as $5^{x} = 7$ .

$5^{x} = 7$

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (5^{x}) = \ln (7)$

Expand $\ln (5^{x})$ by moving $x$ outside the logarithm.

$x \ln (5) = \ln (7)$

Divide each term in $x \ln (5) = \ln (7)$ by $\ln (5)$ and simplify.

Divide each term in $x \ln (5) = \ln (7)$ by $\ln (5)$ .

$\frac{x \ln (5)}{\ln (5)} = \frac{\ln (7)}{\ln (5)}$

Simplify the left side.

Cancel the common factor of $\ln (5)$ .

Cancel the common factor.

$\frac{x \ln (5)}{\ln (5)} = \frac{\ln (7)}{\ln (5)}$

Divide $x$ by $1$ .

$x = \frac{\ln (7)}{\ln (5)}$

Step 8

Substitute $6$ for $u$ in $u = 5^{x}$ .

$6 = 5^{x}$

Step 9

Solve $6 = 5^{x}$ .

Rewrite the equation as $5^{x} = 6$ .

$5^{x} = 6$

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (5^{x}) = \ln (6)$

Expand $\ln (5^{x})$ by moving $x$ outside the logarithm.

$x \ln (5) = \ln (6)$

Divide each term in $x \ln (5) = \ln (6)$ by $\ln (5)$ and simplify.

Divide each term in $x \ln (5) = \ln (6)$ by $\ln (5)$ .

$\frac{x \ln (5)}{\ln (5)} = \frac{\ln (6)}{\ln (5)}$

Simplify the left side.

Cancel the common factor of $\ln (5)$ .

Cancel the common factor.

$\frac{x \ln (5)}{\ln (5)} = \frac{\ln (6)}{\ln (5)}$

Divide $x$ by $1$ .

$x = \frac{\ln (6)}{\ln (5)}$

Step 10

List the solutions that makes the equation true.

$x = \frac{\ln (7)}{\ln (5)}, \frac{\ln (6)}{\ln (5)}$

Step 11

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (7)}{\ln (5)}, \frac{\ln (6)}{\ln (5)}$

Decimal Form:

$x = 1.20906195 \dots, 1.11328275 \dots$

Solve for x 5^(2x)+42=13(5^x)

Related Questions

Support LCMGCF.com by disabling your adblocker.