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:
The solution to the equation is:
- <u>x = 2.5</u> or <u>x = 5/2</u>
Step-by-step explanation:
To find the solution to the equation, you only must operate the equation until you clear the x variable, with the next steps:
And we solve:
- 4x + 28 = 38 (we multiply 4 by x and 7)
- 4x = 38 - 28 (we pass the +28 to subtract to the right side of the equality and operate)
- x = 10 / 4 (we pass the 4 that is multiplying to divide to the right side of the equality)
- <u>x = 2.5</u> (we divide)
As you can see, <u><em>the solution of the equation is </em></u><u><em>x = 2.5</em></u><u><em> or </em></u><u><em>x = 5/2</em></u>
Answer:
c) assets, liabilities, and owner's equity
Step-by-step explanation:
Answer:
b. 23
Step-by-step explanation:
Using factor trees
23 115
∧ ∧
1 23 5 23
So,
23= 23
115= 23
If you circle the common factors in each factor tree, you get 23.
Therefore your answer is B