Find the Roots (Zeros) 25^x-12*5^x+27=0

$25^{x} - 12 \cdot 5^{x} + 27 = 0$

Step 1

Factor the left side of the equation.

Rewrite $25^{x}$ as ${(5^{2})}^{x}$ .

${(5^{2})}^{x} - 12 \cdot 5^{x} + 27 = 0$

Rewrite ${(5^{2})}^{x}$ as ${(5^{x})}^{2}$ .

${(5^{x})}^{2} - 12 \cdot 5^{x} + 27 = 0$

Let $u = 5^{x}$ . Substitute $u$ for all occurrences of $5^{x}$ .

$u^{2} - 12 u + 27 = 0$

Factor $u^{2} - 12 u + 27$ 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 $27$ and whose sum is $- 12$ .

$- 9, - 3$

Write the factored form using these integers.

$(u - 9) (u - 3) = 0$

Replace all occurrences of $u$ with $5^{x}$ .

$(5^{x} - 9) (5^{x} - 3) = 0$

Step 2

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

$5^{x} - 9 = 0$

$5^{x} - 3 = 0$

Step 3

Set $5^{x} - 9$ equal to $0$ and solve for $x$ .

Set $5^{x} - 9$ equal to $0$ .

$5^{x} - 9 = 0$

Solve $5^{x} - 9 = 0$ for $x$ .

Add $9$ to both sides of the equation.

$5^{x} = 9$

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

$\ln (5^{x}) = \ln (9)$

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

$x \ln (5) = \ln (9)$

Divide each term in $x \ln (5) = \ln (9)$ by $\ln (5)$ and simplify.

Divide each term in $x \ln (5) = \ln (9)$ by $\ln (5)$ .

$\frac{x \ln (5)}{\ln (5)} = \frac{\ln (9)}{\ln (5)}$

Simplify the left side.

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

Cancel the common factor.

$\frac{x \ln (5)}{\ln (5)} = \frac{\ln (9)}{\ln (5)}$

Divide $x$ by $1$ .

$x = \frac{\ln (9)}{\ln (5)}$

Step 4

Set $5^{x} - 3$ equal to $0$ and solve for $x$ .

Set $5^{x} - 3$ equal to $0$ .

$5^{x} - 3 = 0$

Solve $5^{x} - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$5^{x} = 3$

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

$\ln (5^{x}) = \ln (3)$

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

$x \ln (5) = \ln (3)$

Divide each term in $x \ln (5) = \ln (3)$ by $\ln (5)$ and simplify.

Divide each term in $x \ln (5) = \ln (3)$ by $\ln (5)$ .

$\frac{x \ln (5)}{\ln (5)} = \frac{\ln (3)}{\ln (5)}$

Simplify the left side.

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

Cancel the common factor.

$\frac{x \ln (5)}{\ln (5)} = \frac{\ln (3)}{\ln (5)}$

Divide $x$ by $1$ .

$x = \frac{\ln (3)}{\ln (5)}$

Step 5

The final solution is all the values that make $(5^{x} - 9) (5^{x} - 3) = 0$ true.

$x = \frac{\ln (9)}{\ln (5)}, \frac{\ln (3)}{\ln (5)}$

Step 6

Find the Roots (Zeros) 25^x-12*5^x+27=0

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