This is going to help you understand
ttps://www.mathsisfun.com/algebra/degree-expression.html
If you meant MG=33
X= 6.8571428571
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:
11. This is counting only odd numbers, or counting by two starting with one.
Answer:
x=28
Step-by-step explanation:
We can use similar triangles and proportions to solve this problem. Put the side of the small triangle over the same side of the larger triangle.
x 42
---------- = ----------
x+10 42+15
Simplify
x 42
---------- = ----------
x+10 57
Using cross products
57x = 42 (x+10)
Distribute
57x = 42x+420
Subtract 42x from each side
57x-42x = 42x-42x +420
15x = 420
Divide each side by 15
15x/15 = 420/15
x=28
