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.
__
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.
__
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:
inferno
Step-by-step explanation:
bb u the reason why hells so hot hell so not ohhhhhhh lallalalala
Answer:
there is a 50% increase of change from 200 to 300. What do you mean by total cost or sale price?
Step-by-step explanation:
The answer is three the ratio as a fraction in simplest form w that equation is 3
The slope is (8-2)/(0-2)=-3
the y intercept is 8
the equation is y=-3x+8
the area to the left is shaded, and the line is solid, so y
-3x+8
