Its option C btw i got help from my tutor with this yesterday
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:
a) 
b) 
Step-by-step explanation:
a) Let the required polynomial be p(x).
We have the relation, 
+ p(x) = 18
i.e. p(x) = 18
i.e. p(x) =
b) Let the required polynomial be q(x).
We have the relation, + q(x) = 0
i.e. q(x) = 0
i.e. q(x) =
Answer:
negative 1 1/3
Step-by-step explanation:
you do y2 minus y1 over x2 minus x1 then you simplify.
