I think the answer is C, but I'm not sure.
1= 15
2= 12
Hope this helps
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.
U=2x+3, x=u/2-3/2
u^2+8u+11=0
u1=(-8+sqrt(64-44))/2=-4+sqrt(5)
u2=-4-sqrt(5)
x1=u1/2-3/2=-7/2+sqrt(5)/2
x2=u2-3/2=-7/2-sqrt(5)/2
Alright so, After rewriting the equation in a standard parabola, you get:
4 x 2 (y-0) =(x - 0)^2
Simplify it to get:
y= 0-p
y= 0-2
When you simplify that, your answer is:
y=-2
There you go folks, just saved you from reading that long paragraph of an answer the other guy has (I don't even think the answer is in there lol, I think he just wanted to help people solve it but still most people come here for answers)
