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:
2,033-993.50= 1039.50/74.25= 14
Step-by-step explanation:
not sure the exact equation but p would be 14 and you'd do something similar to that shown above
Answer:
There are 40 problems on the test
Step-by-step explanation:
Let p = the number of problems on the test
The number of problems on the test times the percent correct is the number correct
p * 85% = 34
p * .85 = 34
Divide each side by .85
p * .85/.85 = 34/.85
p = 40
There are 40 problems on the test
Answer: (2,6) ; 2sqrt10 units
Step-by-step explanation:
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction
PEMDAS = Please Excuse My Dear Aunt Sally
