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:
The shape is not there so i can not solve it
Step-by-step explanation:
Disagree, the fractions are unequal and 10/15=66% while 15/20=75%
Answer:
Doug sold 100 sandwiches in his neighborhood.
Step-by-step explanation:
To solve for percentage you have to first find the percentage's decimal. To find this you have to the move the decimal point up to places. This makes 80 into 0.8. After you do this you will mutiply that but the total amount. So 125 times 0.8. Once you do this you should get 100. This means that Doug sold 100 sandwiches in his neighborhood.
