Solved on Jan 21, 2024

Solve the equation $7 \ln (x) - \ln (x^{2}) = 1$ for the unknown variable $x$ .

STEP 1

Assumptions
1. We are given the equation $7 \ln (x) - \ln \left(x^{2}\right) = 1$ .
2. We assume that $x > 0$ since the natural logarithm is only defined for positive numbers.

STEP 2

Use the properties of logarithms to combine the logarithmic terms on the left side of the equation. The property $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$ can be applied here.
$7 \ln (x) - \ln \left(x^{2}\right) = \ln\left(x^{7}\right) - \ln\left(x^{2}\right)$

STEP 3

Apply the property mentioned in STEP_2 to combine the logarithms into a single logarithm.
$\ln\left(x^{7}\right) - \ln\left(x^{2}\right) = \ln\left(\frac{x^{7}}{x^{2}}\right)$

STEP 4

Simplify the expression inside the logarithm using the exponent rule $\frac{x^{a}}{x^{b}} = x^{a-b}$ .
$\ln\left(\frac{x^{7}}{x^{2}}\right) = \ln\left(x^{7-2}\right) = \ln\left(x^{5}\right)$

STEP 5

Now we have the simplified equation:
$\ln\left(x^{5}\right) = 1$

STEP 6

To solve for $x$ , we need to get rid of the logarithm. We can exponentiate both sides of the equation with base $e$ to do this because $\ln(x) = y$ is equivalent to $e^{y} = x$ .
$e^{\ln\left(x^{5}\right)} = e^{1}$

STEP 7

Simplify the left side using the property that $e^{\ln(a)} = a$ .
$x^{5} = e^{1}$

STEP 8

Since $e^{1} = e$ , we have:
$x^{5} = e$

STEP 9

To solve for $x$ , take the fifth root of both sides of the equation.
$x = \sqrt[5]{e}$

STEP 10

The fifth root of $e$ is the solution for $x$ .
$x = \sqrt[5]{e}$
Therefore, the solution to the equation $7 \ln (x) - \ln \left(x^{2}\right) = 1$ is $x = \sqrt[5]{e}$ .

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Solve the equation $7 \ln (x) - \ln (x^{2}) = 1$ for the unknown variable $x$ .

STEP 1

Assumptions
1. We are given the equation $7 \ln (x) - \ln \left(x^{2}\right) = 1$ .
2. We assume that $x > 0$ since the natural logarithm is only defined for positive numbers.

STEP 2

Use the properties of logarithms to combine the logarithmic terms on the left side of the equation. The property $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$ can be applied here.
$7 \ln (x) - \ln \left(x^{2}\right) = \ln\left(x^{7}\right) - \ln\left(x^{2}\right)$

STEP 3

Apply the property mentioned in STEP_2 to combine the logarithms into a single logarithm.
$\ln\left(x^{7}\right) - \ln\left(x^{2}\right) = \ln\left(\frac{x^{7}}{x^{2}}\right)$

STEP 4

Simplify the expression inside the logarithm using the exponent rule $\frac{x^{a}}{x^{b}} = x^{a-b}$ .
$\ln\left(\frac{x^{7}}{x^{2}}\right) = \ln\left(x^{7-2}\right) = \ln\left(x^{5}\right)$

STEP 5

Now we have the simplified equation:
$\ln\left(x^{5}\right) = 1$

STEP 6

To solve for $x$ , we need to get rid of the logarithm. We can exponentiate both sides of the equation with base $e$ to do this because $\ln(x) = y$ is equivalent to $e^{y} = x$ .
$e^{\ln\left(x^{5}\right)} = e^{1}$

STEP 7

Simplify the left side using the property that $e^{\ln(a)} = a$ .
$x^{5} = e^{1}$

STEP 8

Since $e^{1} = e$ , we have:
$x^{5} = e$

STEP 9

To solve for $x$ , take the fifth root of both sides of the equation.
$x = \sqrt[5]{e}$

STEP 10

The fifth root of $e$ is the solution for $x$ .
$x = \sqrt[5]{e}$
Therefore, the solution to the equation $7 \ln (x) - \ln \left(x^{2}\right) = 1$ is $x = \sqrt[5]{e}$ .

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Solve the equation 7ln⁡(x)−ln⁡(x2)=17 \ln (x) - \ln (x^{2}) = 17ln(x)−ln(x2)=1 for the unknown variable xxx.

STEP 1

Assumptions1. We are given the equation 7ln⁡(x)−ln⁡(x2)=17 \ln (x) - \ln \left(x^{2}\right) = 17ln(x)−ln(x2)=1.2. We assume that x>0x > 0x>0 since the natural logarithm is only defined for positive numbers.

STEP 2

STEP 3

Apply the property mentioned in STEP_2 to combine the logarithms into a single logarithm.ln⁡(x7)−ln⁡(x2)=ln⁡(x7x2)\ln\left(x^{7}\right) - \ln\left(x^{2}\right) = \ln\left(\frac{x^{7}}{x^{2}}\right)ln(x7)−ln(x2)=ln(x2x7​)

STEP 4

Simplify the expression inside the logarithm using the exponent rule xaxb=xa−b\frac{x^{a}}{x^{b}} = x^{a-b}xbxa​=xa−b.ln⁡(x7x2)=ln⁡(x7−2)=ln⁡(x5)\ln\left(\frac{x^{7}}{x^{2}}\right) = \ln\left(x^{7-2}\right) = \ln\left(x^{5}\right)ln(x2x7​)=ln(x7−2)=ln(x5)

STEP 5

Now we have the simplified equation:ln⁡(x5)=1\ln\left(x^{5}\right) = 1ln(x5)=1

STEP 6

To solve for xxx, we need to get rid of the logarithm. We can exponentiate both sides of the equation with base eee to do this because ln⁡(x)=y\ln(x) = yln(x)=y is equivalent to ey=xe^{y} = xey=x.eln⁡(x5)=e1e^{\ln\left(x^{5}\right)} = e^{1}eln(x5)=e1

STEP 7

Simplify the left side using the property that eln⁡(a)=ae^{\ln(a)} = aeln(a)=a.x5=e1x^{5} = e^{1}x5=e1

STEP 8

Since e1=ee^{1} = ee1=e, we have:x5=ex^{5} = ex5=e

STEP 9

To solve for xxx, take the fifth root of both sides of the equation.x=e5x = \sqrt[5]{e}x=5e​

STEP 10

The fifth root of eee is the solution for xxx.x=e5x = \sqrt[5]{e}x=5e​Therefore, the solution to the equation 7ln⁡(x)−ln⁡(x2)=17 \ln (x) - \ln \left(x^{2}\right) = 17ln(x)−ln(x2)=1 is x=e5x = \sqrt[5]{e}x=5e​.

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Solve the equation $7 \ln (x) - \ln (x^{2}) = 1$ for the unknown variable $x$ .

Assumptions
1. We are given the equation $7 \ln (x) - \ln \left(x^{2}\right) = 1$ .
2. We assume that $x > 0$ since the natural logarithm is only defined for positive numbers.

Apply the property mentioned in STEP_2 to combine the logarithms into a single logarithm.
$\ln\left(x^{7}\right) - \ln\left(x^{2}\right) = \ln\left(\frac{x^{7}}{x^{2}}\right)$

Simplify the expression inside the logarithm using the exponent rule $\frac{x^{a}}{x^{b}} = x^{a-b}$ .
$\ln\left(\frac{x^{7}}{x^{2}}\right) = \ln\left(x^{7-2}\right) = \ln\left(x^{5}\right)$

Now we have the simplified equation:
$\ln\left(x^{5}\right) = 1$

To solve for $x$ , we need to get rid of the logarithm. We can exponentiate both sides of the equation with base $e$ to do this because $\ln(x) = y$ is equivalent to $e^{y} = x$ .
$e^{\ln\left(x^{5}\right)} = e^{1}$

Simplify the left side using the property that $e^{\ln(a)} = a$ .
$x^{5} = e^{1}$

Since $e^{1} = e$ , we have:
$x^{5} = e$

To solve for $x$ , take the fifth root of both sides of the equation.
$x = \sqrt[5]{e}$

The fifth root of $e$ is the solution for $x$ .
$x = \sqrt[5]{e}$
Therefore, the solution to the equation $7 \ln (x) - \ln \left(x^{2}\right) = 1$ is $x = \sqrt[5]{e}$ .