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.
3x9=27
21/3 = 7
so
7+27x9+5
27x9=243
243+7+5
answer is 255
Answer: 3.36
Step-by-step explanation: 75minutes / 28= 2.678571428571429
9/2.678571428571429=3.36 times
(X+4)(X-4) is the LCD because you just have to look at the denominators and see what factors or values are not already part of your LCD.
Step-by-step explanation:
If you use synthetic division, you get,
Which is,
Answered by GAUTHMATH
