Solve the exponential equation. Express the solution in terms of logarithms.

7^(4x-2) = 6^(x-2)

Question

Solve the exponential equation.  Express the solution in terms of logarithms.  

7^(4x-2) = 6^(x-2)

campbell_st · Answer

take the log of both sides... doesn't matter if its base e or base 10

$$\log (7^{4x - 2} = \log(6^{x - 2})$$

then by using log laws

$$(4x -2)\log(7) = (x -1) \log(6)$$

expand anc collect like terms 

4xlog(7) - x log(6) = 2log(7) - log(6)

x(4log(7) - log(6)) = 2log(7) - log(6)

$$x = \frac{2\log(7) - \log(6)}{2\log(7) - \log(6)}$$

which can be simplified further to 

$$x = \frac{\log(\frac{7^4}{6})}{\log(\frac{7^2}{6})}$$

anonymous · Answer

perfect thank you.  awesome step by step process

anonymous · Answer

I believe that this problem was solved in reverse. The answer should be the same as given only with the numerator and denominator swapped. so Log(7^4/6) should be in the denominator.

anonymous · Answer

$$x = \log(7^2/6)/\log(7^4/6)$$