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: It would of been $339.25
Step-by-step explanation: If you do the math 299 divided by 45 you will get 40 so now add 40 to 299 you will get $339 minus 50 cents will be 25 so $339.25 would be the original price of the bike.
The equation of a horizontal line is
y
=
c
for some constant
c
Since we are told that the line passes through
(
x
,
y
)
=
(
−
4
,
6
)
Then for at least one point on the line
y
=
6
But for a horizontal line this value is a constant for all points on the line.
So the equation for all points on the line is
y
=
6
Answer:
the area for the square is 126.56cm²:
