Solve for x (16/25)^(x^2+x)=1

${(\frac{16}{25})}^{x^{2} + x} = 1$

Step 1

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

$\ln ({(\frac{16}{25})}^{x^{2} + x}) = \ln (1)$

Step 2

Expand the left side.

Expand $\ln ({(\frac{16}{25})}^{x^{2} + x})$ by moving $x^{2} + x$ outside the logarithm.

$(x^{2} + x) \ln (\frac{16}{25}) = \ln (1)$

Rewrite $\ln (\frac{16}{25})$ as $\ln (16) - \ln (25)$ .

$(x^{2} + x) (\ln (16) - \ln (25)) = \ln (1)$

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

$(x^{2} + x) (\ln (16) - \ln (5^{2})) = \ln (1)$

Expand $\ln (5^{2})$ by moving $2$ outside the logarithm.

$(x^{2} + x) (\ln (16) - (2 \ln (5))) = \ln (1)$

Multiply $2$ by $- 1$ .

$(x^{2} + x) (\ln (16) - 2 \ln (5)) = \ln (1)$

Step 3

Simplify the left side.

Simplify $(x^{2} + x) (\ln (16) - 2 \ln (5))$ .

Expand $(x^{2} + x) (\ln (16) - 2 \ln (5))$ using the FOIL Method.

Apply the distributive property.

$x^{2} (\ln (16) - 2 \ln (5)) + x (\ln (16) - 2 \ln (5)) = \ln (1)$

Apply the distributive property.

$x^{2} \ln (16) + x^{2} (- 2 \ln (5)) + x (\ln (16) - 2 \ln (5)) = \ln (1)$

Apply the distributive property.

$x^{2} \ln (16) + x^{2} (- 2 \ln (5)) + x \ln (16) + x (- 2 \ln (5)) = \ln (1)$

Reorder factors in $x^{2} \ln (16) + x^{2} (- 2 \ln (5)) + x \ln (16) + x (- 2 \ln (5))$ .

$x^{2} \ln (16) - 2 x^{2} \ln (5) + x \ln (16) - 2 x \ln (5) = \ln (1)$

Step 4

Simplify the right side.

The natural logarithm of $1$ is $0$ .

$x^{2} \ln (16) - 2 x^{2} \ln (5) + x \ln (16) - 2 x \ln (5) = 0$

Step 5

Factor $x$ out of $x^{2} \ln (16) - 2 x^{2} \ln (5) + x \ln (16) - 2 x \ln (5)$ .

Factor $x$ out of $x^{2} \ln (16)$ .

$x (x \ln (16)) - 2 x^{2} \ln (5) + x \ln (16) - 2 x \ln (5) = 0$

Factor $x$ out of $- 2 x^{2} \ln (5)$ .

$x (x \ln (16)) + x (- 2 x \ln (5)) + x \ln (16) - 2 x \ln (5) = 0$

Factor $x$ out of $x \ln (16)$ .

$x (x (\ln (16))) + x (- 2 x \ln (5)) + x (\ln (16)) - 2 x \ln (5) = 0$

Factor $x$ out of $- 2 x \ln (5)$ .

$x (x (\ln (16))) + x (- 2 x \ln (5)) + x (\ln (16)) + x (- 2 \ln (5)) = 0$

Factor $x$ out of $x (x \ln (16)) + x (- 2 x \ln (5))$ .

$x (x (\ln (16)) - 2 x \ln (5)) + x (\ln (16)) + x (- 2 \ln (5)) = 0$

Factor $x$ out of $x (x (\ln (16)) - 2 x \ln (5)) + x (\ln (16))$ .

$x (x \ln (16) - 2 x \ln (5) + \ln (16)) + x (- 2 \ln (5)) = 0$

Factor $x$ out of $x (x \ln (16) - 2 x \ln (5) + \ln (16)) + x (- 2 \ln (5))$ .

$x (x \ln (16) - 2 x \ln (5) + \ln (16) - 2 \ln (5)) = 0$

Step 6

Factor out the greatest common factor (GCF) from each group.

$x (x (\ln (16) - 2 \ln (5)) + 1 (\ln (16) - 2 \ln (5))) = 0$

Step 7

Factor.

Factor the polynomial by factoring out the greatest common factor, $\ln (16) - 2 \ln (5)$ .

$x ((\ln (16) - 2 \ln (5)) (x + 1)) = 0$

Remove unnecessary parentheses.

$x (\ln (16) - 2 \ln (5)) (x + 1) = 0$

Step 8

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

$x = 0$

$x + 1 = 0$

Step 9

Set $x$ equal to $0$ .

$x = 0$

Step 10

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 11

The final solution is all the values that make $x (\ln (16) - 2 \ln (5)) (x + 1) = 0$ true.

$x = 0, - 1$

Solve for x (16/25)^(x^2+x)=1

Related Questions

Support LCMGCF.com by disabling your adblocker.