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:
Use the Pythagorean theorem for right triangles to solve this
a^2 + b^2 = c^2 where 'a' and 'b' are the legs of the triangle and 'c' is the hypotenuse
so:
8^2 + b^2 = 17^2 solve for the second leg, b
64 + b^2 = 289
b^2 = 289 - 64= 225
b = 15
Y² + 8y + 15
since 8 and fifteen are positive, that means your factors are going to be positive
(y + )(y + )
What are the factors of 15 that add up to 8?
An inverse operation reverses the effect of another operation. Answer is C
Answer:
Lol ..... I don't know the answer :D :P
Step-by-step explanation:
L.H.S x 0 = 0
R.H.S x 0 = 0
L.H.S = R.H.S
Hence , proved .....
