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.
__
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.
__
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.
Convert both fractions to get a LCD, then subtract
Answer is 3. 1/24
Answer:
640
Step-by-step explanation:
there is 1 gallon for every 128 ounces
so 1 gal= 128z
to find 5gal= ?oz
you multiply 128 by 5 which gives u 640
which gives u 640 oz
hope this helps
Answer:
<u>72 mm total and 10.29 mm average.</u>
Step-by-step explanation:
16 mm
14 mm
6 mm
3 mm
18 mm
5 mm
<u>10</u> <u>mm </u>
<u>72</u> mm total length
Divide by 7: 72/7 = <u>10.29</u> mm
Answer:
I would love to help!!
Step-by-step explanation:
