Answer:
x = -7 or x = 2/3
Step-by-step explanation:
I'm assuming you meant solve for x when f(x) = 0.
f(x) = 3x(x + 7) - 2(x + 7)
0 = (x + 7)(3x - 2) -- Both terms have a common factor of (x + 7) so we can group them
x + 7 = 0 or 3x - 2 = 0 -- Use ZPP
x = -7 or x = 2/3 -- Solve
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.
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C) you conclude that less than half of the cars on the dealership’s lot cost more than$34,330
