Solve for x 27(3/5)^(x+1)=125

$27 {(\frac{3}{5})}^{x + 1} = 125$

Step 1

Divide each term in $27 {(\frac{3}{5})}^{x + 1} = 125$ by $27$ and simplify.

Divide each term in $27 {(\frac{3}{5})}^{x + 1} = 125$ by $27$ .

$\frac{27 {(\frac{3}{5})}^{x + 1}}{27} = \frac{125}{27}$

Simplify the left side.

Cancel the common factor of $27$ .

Cancel the common factor.

$\frac{27 {(\frac{3}{5})}^{x + 1}}{27} = \frac{125}{27}$

Divide ${(\frac{3}{5})}^{x + 1}$ by $1$ .

${(\frac{3}{5})}^{x + 1} = \frac{125}{27}$

Apply the product rule to $\frac{3}{5}$ .

$\frac{3^{x + 1}}{5^{x + 1}} = \frac{125}{27}$

Step 2

Multiply both sides by $5^{x + 1}$ .

$\frac{3^{x + 1}}{5^{x + 1}} \cdot 5^{x + 1} = \frac{125}{27} \cdot 5^{x + 1}$

Step 3

Simplify.

Simplify the left side.

Cancel the common factor of $5^{x + 1}$ .

Cancel the common factor.

$\frac{3^{x + 1}}{5^{x + 1}} \cdot 5^{x + 1} = \frac{125}{27} \cdot 5^{x + 1}$

Rewrite the expression.

$3^{x + 1} = \frac{125}{27} \cdot 5^{x + 1}$

Simplify the right side.

Multiply $\frac{125}{27} \cdot 5^{x + 1}$ .

Combine $\frac{125}{27}$ and $5^{x + 1}$ .

$3^{x + 1} = \frac{125 \cdot 5^{x + 1}}{27}$

Rewrite $125$ as $5^{3}$ .

$3^{x + 1} = \frac{5^{3} \cdot 5^{x + 1}}{27}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$3^{x + 1} = \frac{5^{3 + x + 1}}{27}$

Add $3$ and $1$ .

$3^{x + 1} = \frac{5^{x + 4}}{27}$

Step 4

Solve for $x$ .

Take the log of both sides of the equation.

$\ln (3^{x + 1}) = \ln (\frac{5^{x + 4}}{27})$

Expand $\ln (3^{x + 1})$ by moving $x + 1$ outside the logarithm.

$(x + 1) \ln (3) = \ln (\frac{5^{x + 4}}{27})$

Rewrite $\ln (\frac{5^{x + 4}}{27})$ as $\ln (5^{x + 4}) - \ln (27)$ .

$(x + 1) \ln (3) = \ln (5^{x + 4}) - \ln (27)$

Expand $\ln (5^{x + 4})$ by moving $x + 4$ outside the logarithm.

$(x + 1) \ln (3) = (x + 4) \ln (5) - \ln (27)$

Rewrite $\ln (27)$ as $\ln (3^{3})$ .

$(x + 1) \ln (3) = (x + 4) \ln (5) - \ln (3^{3})$

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

$(x + 1) \ln (3) = (x + 4) \ln (5) - (3 \ln (3))$

Multiply $3$ by $- 1$ .

$(x + 1) \ln (3) = (x + 4) \ln (5) - 3 \ln (3)$

Solve the equation for $x$ .

Simplify the left side.

Simplify $(x + 1) \ln (3)$ .

Apply the distributive property.

$x \ln (3) + 1 \ln (3) = (x + 4) \ln (5) - 3 \ln (3)$

Multiply $\ln (3)$ by $1$ .

$x \ln (3) + \ln (3) = (x + 4) \ln (5) - 3 \ln (3)$

Simplify the right side.

Simplify $(x + 4) \ln (5) - 3 \ln (3)$ .

Simplify each term.

Apply the distributive property.

$x \ln (3) + \ln (3) = x \ln (5) + 4 \ln (5) - 3 \ln (3)$

Simplify $4 \ln (5)$ by moving $4$ inside the logarithm.

$x \ln (3) + \ln (3) = x \ln (5) + \ln (5^{4}) - 3 \ln (3)$

Raise $5$ to the power of $4$ .

$x \ln (3) + \ln (3) = x \ln (5) + \ln (625) - 3 \ln (3)$

Simplify $- 3 \ln (3)$ by moving $3$ inside the logarithm.

$x \ln (3) + \ln (3) = x \ln (5) + \ln (625) - \ln (3^{3})$

Raise $3$ to the power of $3$ .

$x \ln (3) + \ln (3) = x \ln (5) + \ln (625) - \ln (27)$

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$x \ln (3) + \ln (3) = x \ln (5) + \ln (\frac{625}{27})$

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

$x \ln (3) + \ln (3) - x \ln (5) - \ln (\frac{625}{27}) = 0$

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$x \ln (3) - x \ln (5) + \ln (\frac{3}{\frac{625}{27}}) = 0$

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$x \ln (3) - x \ln (5) + \ln (3 (\frac{27}{625})) = 0$

Multiply $3 (\frac{27}{625})$ .

Combine $3$ and $\frac{27}{625}$ .

$x \ln (3) - x \ln (5) + \ln (\frac{3 \cdot 27}{625}) = 0$

Multiply $3$ by $27$ .

$x \ln (3) - x \ln (5) + \ln (\frac{81}{625}) = 0$

Subtract $\ln (\frac{81}{625})$ from both sides of the equation.

$x \ln (3) - x \ln (5) = - \ln (\frac{81}{625})$

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

Factor $x$ out of $x \ln (3)$ .

$x (\ln (3)) - x \ln (5) = - \ln (\frac{81}{625})$

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

$x (\ln (3)) + x (- 1 \ln (5)) = - \ln (\frac{81}{625})$

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

$x (\ln (3) - 1 \ln (5)) = - \ln (\frac{81}{625})$

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

$x (\ln (3) - \ln (5)) = - \ln (\frac{81}{625})$

Divide each term in $x (\ln (3) - \ln (5)) = - \ln (\frac{81}{625})$ by $\ln (3) - \ln (5)$ and simplify.

Divide each term in $x (\ln (3) - \ln (5)) = - \ln (\frac{81}{625})$ by $\ln (3) - \ln (5)$ .

$\frac{x (\ln (3) - \ln (5))}{\ln (3) - \ln (5)} = \frac{- \ln (\frac{81}{625})}{\ln (3) - \ln (5)}$

Simplify the left side.

Cancel the common factor of $\ln (3) - \ln (5)$ .

Cancel the common factor.

$\frac{x (\ln (3) - \ln (5))}{\ln (3) - \ln (5)} = \frac{- \ln (\frac{81}{625})}{\ln (3) - \ln (5)}$

Divide $x$ by $1$ .

$x = \frac{- \ln (\frac{81}{625})}{\ln (3) - \ln (5)}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{\ln (\frac{81}{625})}{\ln (3) - \ln (5)}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{\ln (\frac{81}{625})}{\ln (3) - \ln (5)}$

Decimal Form:

$x = - 4$

Solve for x 27(3/5)^(x+1)=125

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