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.
Answer:
207°
Step-by-step explanation:
The answer:
8 times
Explanation:
If you divide 168 by 20, you get 8.4, and the whole number is 8 so that’s the answer.
Answer:108
Step-by-step explanation:
2 because once you double it, it becomes 4. Once you add 4, you get 8. When you subtract your number, which is 2, you get 6. Then when you subtract 3, you get 3. 3 is your number (2) plus 1
