(-2,8) This is a answer:)
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:
-3/2
Step-by-step explanation:
Using the slope formula, we can find the slope of the line shown
m = ( y2-y1)/(x2-x1)
= ( -1-3)/(-3-3)
= -4/-6
= 2/3
A line that is perpendicular is the negative reciprocal
-1/ (2/3) = -3/2
Answer:
The Answer is 25
Step-by-step explanation:
