Answer:
use logarithms
Step-by-step explanation:
Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.
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You will note that this approach works well enough for ...
a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents
(x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs
but doesn't do anything to help you solve ...
x +3 = b^(x -6)
There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.
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Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.
In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.
If the first roll is a 4, the possibilities to add less than 8 are that the second cube roll is 1, 2 or 3.
These are 3 different events out of 6 possilbe ones, which is 3 / 6 = 1 / 2; i.e. half of times.
Then the answer is the first option: 1 / 2
How many minutes did it take him to cover 1 mile? 5 minutes.
-11.1666667 should be the answer
