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.
Step-by-step explanation:
sin(2x) + cos(3x)
Use double angle formula sin(2x) = 2 sin x cos x.
Use triple angle formula cos(3x) = 4 cos³x − 3 cos x.
Substitute:
2 sin x cos x + 4 cos³x − 3 cos x
Answer:
1.5 yards
Step-by-step explanation:
just looked it up
Answer:
Step-by-step explanation:
30% of a dollar...
0.30(1) = 0.30.....so 30% of a dollar is 30 cents
1 quarter and 1 nickel = 0.30
3 dimes = 0.30
2 dimes and 2 nickels = 0.30
6 nickels = 0.30
3 nickels, 1 dime, 5 pennies = 0.30
10 pennies, 2 nickels, a dime = 0.30
30 pennies = 0.30
take ur pick :)
Answer:
25/784
Step-by-step explanation:
