Solve for x 5^(7x+1)=7^(x-5)

Problem

5(7*x+1)=7(x−5)

Solution

1. Take the natural logarithm of both sides of the equation to move the variables out of the exponents.

ln(5(7*x+1))=ln(7(x−5))

2. Apply the power rule for logarithms, ln(ab)=b*ln(a) to both sides.

(7*x+1)*ln(5)=(x−5)*ln(7)

3. Distribute the logarithmic constants into the binomials.

7*x*ln(5)+ln(5)=x*ln(7)−5*ln(7)

4. Isolate the terms containing x on one side of the equation and the constants on the other.

7*x*ln(5)−x*ln(7)=−5*ln(7)−ln(5)

5. Factor out the common factor x from the left side.

x*(7*ln(5)−ln(7))=−5*ln(7)−ln(5)

6. Divide by the coefficient of x to solve for the variable.

x=(−5*ln(7)−ln(5))/(7*ln(5)−ln(7))

7. Simplify the expression by factoring out a negative sign if desired.

x=−(5*ln(7)+ln(5))/(7*ln(5)−ln(7))

Final Answer

x=(−5*ln(7)−ln(5))/(7*ln(5)−ln(7))

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