Identify the Zeros and Their Multiplicities f(x)=e^(7x-3)-5^(x-6)

$f (x) = e^{7 x - 3} - 5^{x - 6}$

Step 1

Set $e^{7 x - 3} - 5^{x - 6}$ equal to $0$ .

$e^{7 x - 3} - 5^{x - 6} = 0$

Step 2

Solve for $x$ .

Move $- 5^{x - 6}$ to the right side of the equation by adding it to both sides.

$e^{7 x - 3} = 5^{x - 6}$

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

$\ln (e^{7 x - 3}) = \ln (5^{x - 6})$

Expand the left side.

Expand $\ln (e^{7 x - 3})$ by moving $7 x - 3$ outside the logarithm.

$(7 x - 3) \ln (e) = \ln (5^{x - 6})$

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

$(7 x - 3) \cdot 1 = \ln (5^{x - 6})$

Multiply $7 x - 3$ by $1$ .

$7 x - 3 = \ln (5^{x - 6})$

Expand $\ln (5^{x - 6})$ by moving $x - 6$ outside the logarithm.

$7 x - 3 = (x - 6) \ln (5)$

Simplify the right side.

Apply the distributive property.

$7 x - 3 = x \ln (5) - 6 \ln (5)$

Move all the terms containing a logarithm to the left side of the equation.

$- x \ln (5) + 6 \ln (5) = - 7 x + 3$

Add $7 x$ to both sides of the equation.

$- x \ln (5) + 6 \ln (5) + 7 x = 3$

Subtract $6 \ln (5)$ from both sides of the equation.

$- x \ln (5) + 7 x = 3 - 6 \ln (5)$

Factor $x$ out of $- x \ln (5) + 7 x$ .

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

$x (- 1 \ln (5)) + 7 x = 3 - 6 \ln (5)$

Factor $x$ out of $7 x$ .

$x (- 1 \ln (5)) + x \cdot 7 = 3 - 6 \ln (5)$

Factor $x$ out of $x (- 1 \ln (5)) + x \cdot 7$ .

$x (- 1 \ln (5) + 7) = 3 - 6 \ln (5)$

Rewrite $- 1 \ln (5)$ as $- \ln (5)$ .

$x (- \ln (5) + 7) = 3 - 6 \ln (5)$

Divide each term in $x (- \ln (5) + 7) = 3 - 6 \ln (5)$ by $- \ln (5) + 7$ and simplify.

Divide each term in $x (- \ln (5) + 7) = 3 - 6 \ln (5)$ by $- \ln (5) + 7$ .

$\frac{x (- \ln (5) + 7)}{- \ln (5) + 7} = \frac{3}{- \ln (5) + 7} + \frac{- 6 \ln (5)}{- \ln (5) + 7}$

Simplify the left side.

Cancel the common factor of $- \ln (5) + 7$ .

Cancel the common factor.

$\frac{x (- \ln (5) + 7)}{- \ln (5) + 7} = \frac{3}{- \ln (5) + 7} + \frac{- 6 \ln (5)}{- \ln (5) + 7}$

Divide $x$ by $1$ .

$x = \frac{3}{- \ln (5) + 7} + \frac{- 6 \ln (5)}{- \ln (5) + 7}$

Simplify the right side.

Combine the numerators over the common denominator.

$x = \frac{3 - 6 \ln (5)}{- \ln (5) + 7}$

Factor $3$ out of $3 - 6 \ln (5)$ .

Factor $3$ out of $3$ .

$x = \frac{3 (1) - 6 \ln (5)}{- \ln (5) + 7}$

Factor $3$ out of $- 6 \ln (5)$ .

$x = \frac{3 (1) + 3 (- 2 \ln (5))}{- \ln (5) + 7}$

Factor $3$ out of $3 (1) + 3 (- 2 \ln (5))$ .

$x = \frac{3 (1 - 2 \ln (5))}{- \ln (5) + 7}$

Factor $- 1$ out of $- \ln (5)$ .

$x = \frac{3 (1 - 2 \ln (5))}{- \ln (5) + 7}$

Rewrite $7$ as $- 1 (- 7)$ .

$x = \frac{3 (1 - 2 \ln (5))}{- \ln (5) - 1 (- 7)}$

Factor $- 1$ out of $- \ln (5) - 1 (- 7)$ .

$x = \frac{3 (1 - 2 \ln (5))}{- (\ln (5) - 7)}$

Rewrite negatives.

Rewrite $- (\ln (5) - 7)$ as $- 1 (\ln (5) - 7)$ .

$x = \frac{3 (1 - 2 \ln (5))}{- 1 (\ln (5) - 7)}$

Move the negative in front of the fraction.

$x = - \frac{3 (1 - 2 \ln (5))}{\ln (5) - 7}$

$x = - \frac{3 (1 - 2 \ln (5))}{\ln (5) - 7}$ (Multiplicity of $1$ )

Step 3

Identify the Zeros and Their Multiplicities f(x)=e^(7x-3)-5^(x-6)

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