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.
Yea I'm pretty sure it's all right not 100% positive
The first equation shows C and D, and the second shows C and B. The overlap will be at C, so thats the answer.
7/8 / 11/12. First get the reciprocal of 11/12 (which is 12/11). Then multiply by the reciprocal. 7/8 * 12/11 = 84/88
don't forget to simplify
21/22
Remember
all angles add to 180
x+y+z=180
one is 2 times as another
x=2y
measure of 3rd angle (z) is 20 more than smalllest
since y has to be doubled to get x, y is smallest
z is 20 more than y
z=20+y
we have
x=2y
z=20+y
sub those
x+y+z=180
2y+y+20+y=180
4y+20=180
minus 20 both sides
4y=160
divide 4
y=40
sub back
x=2y
x=2(40)
x=80
z=20+y
z=20+40
z=60
the angles are
80 degrees, 40 degrees, 60 degrees
