Question

6^(6x)=7^(x-7)
How would I find the answer in base 10 or base e logarithms?

Answer 1

To solve this equation using logarithms, we can take the logarithm of both sides of the equation.

Using the change of base formula, we can take the logarithm of both sides of the equation with any base we like. Let's use the base 10 logarithm:

log base 10 (6^(6x)) = log base 10 (7^(x-7))

Now we can use the exponent rule of logarithms, which tells us that the exponent of a power can be moved out in front of a logarithm.

6x log base 10 (6) = (x-7) log base 10 (7)

6x log(6) = (x-7) log(7)

Now we can solve for x:

6x log(6) = x log(7) - 7 log(7)

6x log(6) - x log(7) = -7 log(7)

x (6 log(6) - log(7)) = -7 log(7)

x = (-7 log(7)) / (6 log(6) - log(7))

Using the change of base formula again, we can switch from the base 10 logarithm to the natural logarithm (ln):

x = (-7 ln(7)) / (6 ln(6) - ln(7))

This is the solution to the equation in terms of base e logarithms.

Alternatively, to find the solution in base 10 logarithms, you can use the formula:

log base b (a^c) = c log base b (a)

Using this formula, we can rewrite the original equation as:

6x log base 10 (6) = (x - 7) log base 10 (7)

Distributing the logarithm on the right-hand side, we get:

6x log base 10 (6) = x log base 10 (7) - 7 log base 10 (7)

Bringing all the x terms to one side and factoring out x, we get:

x (6 log base 10 (6) - log base 10 (7)) = -7 log base 10 (7)

Dividing both sides by (6 log base 10 (6) - log base 10 (7)), we get:

x = -7 log base 10 (7) / (6 log base 10 (6) - log base 10 (7))

This is the solution to the equation in terms of base 10 logarithms.

Answer 2

See below

Explanation:

Both ways are the same, just different bases:

`\displaystyle 6^(6x)=7^(x-7)\\6x\ln(6)=(x-7)\ln(7)\\6x\ln(6)=x\ln(7)-7\ln(7)\\6x\ln(6)-x\ln(7)=-7\ln(7)\\\\x(6\ln(6)-\ln(7))=-7\ln(7)\\\\x=(-7\ln(7))/(6\ln(6)-\ln(7))\,\, \text{OR}\,\,x=(-7\log(7))/(6\log(6)-\log(7))`

You can use either one as your solution, but like I said, base is irrelevant when doing these kinds of problems (I did ln because I'm more used to it).