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.
The area of the ceiling:
A = 14² = 196 ft²
We need plaster to cover 196 ft².
196 : 90 = 2.17778
Answer: We have to buy 3 pails of plaster.
Answer:
7i=4.5
Step-by-step explanation:
Hope this helps
For a linear equation like this there will always be infinitely many solutions if you let x and y be any real number. So you can just pick any x and find the corresponding y. Take for exmaple x=1 then we get
-1+2y=8
2y=9
y=9/2=4.5 so an ordered pair that satifies your equation is (x,y)=(1,9/2).
Now you can go on and find more.
