Answer:
errday i b countin up countin up da blues count away sometimes we always lose
Step-by-step explanation:
juice wrld
Answer:
6.57, 6.38, 6.125, 6.057, 6.03, 6.01
Answers
1) 4
2) 0, 2, -10
3) 0, 5, -2
Step-by-step explanation:
1) x³ - 64 = 0
x³ = 64
x = 4
2) x³(x² + 8x - 20) = 0
x³(x² + 10x - 2x - 20) = 0
x³(x(x + 10) - 2(x + 10)) = 0
x³(x + 10)(x - 2) = 0
x = 0, 2, -10
3) x³ - 3x² - 10x = 0
x(x² - 3x - 10) = 0
x(x² - 5x + 2x - 10) = 0
x(x(x - 5) + 2(x - 5)) = 0
x(x - 5)(x + 2) = 0
x = 0, 5, -2
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.