It’s probably most definitely the third one but I really don’t know.
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.
Givens
Work week = 40 hours
Paid 7.50 an hour.
Sells 6300 worth of goods.
Equation
Take home pay = Sales * 5% + 7.50 * number of hours.
Sub and Solve
Take home pay = 6300 * 5% + 7.50 * 40
Take home pay = 6300 * 5/100 + 7.50 * 40
Take home pay = 315 + 300
Take home pay = 615 dollars.
Answer 615 <<<<
Answer:
i think they are y and (x,y)
Answer:
96
Step-by-step explanation:
60/100 x 60 = 96
