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:
a(a + 3b)
Step-by-step explanation:
a² + 2ab + ab
a² + 3ab
a(a + 3b)
Ratio and proportion
real:shadow
so the person is 1.8m but the shadow is 4m
so
real:shadow=1.8:4
so
1.8:4=?:14
1.8/4=?/14
times both sides by 14
25.2/4=?
6.3=?
the tree is 6.3 m tall
Answer:
y=mx+b
also if you ask me how I got that is because you need to listen carefully to the math teacher word so you can remember the answer
