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To solve an equation like this you start by finding the logarithm of each side. You can use any base for the logarithm but the simplest, exact answer will be found using the same base logarithm as the base of the exponent. In this case, base 6:
Now we can use a property of logarithms, , to move the exponent of the argument out in front. Being able to use this property to get the variable out of the exponent is the whole reason we use logarithms on equations where the variable is in an exponent. Using this property on your equation we get:
Since (which is why base 6 logarithms give us the simplest answer) this simplifies to:
If you want a decimal approximation of the answer, we will need calculators and logarithms our calculator can find (usually base 10 or base e (ln)). There is a base conversion formula for logarithms: . We can use this to convert our base 6 logarithm to base 10:
or to base e logarithms:
Both will result in the same approximation for the answer. (Because even calculators will use rounded-off decimal approximations for the logarithms of 40 and 6 the final answer you get from base 10 logarithms may be a very tiny bit different from the answer you get using base e logarithms.)
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