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:
EXCEPT FOR HOUSING, NEW YORK SEEMS TO BE PROMINANT IN TERMS OF LIVING.
Step-by-step explanation:
THEREFORE,THE PRICE OF GAS IN NEW YORK WOULD BE MORE THANT THAT OF THE PRICE IN LA.IT WOULD BE SURLEY MORE THAN $3.
AS THE NATIONAL AVERAGE TENDS TO BE AROUND $2.75.
Answer:
Optioin 4
Step-by-step explanation:
12/8 = 15/(x-4)
12x - 48 = 120
12x = 168
x = 14
I think the final answer would be 2x/3
The correct answers are 5, 10, and 15. I'm doing the same subject and I got a 100 on a question similar. Good luck!
