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:
x = 181 and y = 97
Step-by-step explanation:
let called the first number is x
the second number would be called y
We are given that:
x + y = 278 (1)
x = y + 84 (2)
Let change x in (2) into (1):
y + 84 + y = 278
2y + 84 = 278
Subtract 84 from both side, we got:
2y + 84 - 84 = 278 - 84
2y + 0 = 194
Divide both side by 2, we got:
2y / 2 = 194 / 2
y = 97
Because y = 97 and x + y = 278 so x would equal:
x + 97 = 278
Subtract 97 from both side, we got:
x + 97 - 97 = 278 - 97
x + 0 = 181
x = 181 and y = 97
Hope this helped :3
Answer:
option B
polynomial cannot have negative power
hope it helps
