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:
okie
Step-by-step explanation:
Answer:
f(x)=(1.023) ⋅ 3^x Growth
f(x)=3 ⋅ (0.072)^x Decay
f(x)=4 ⋅ (0.035)^x Decay
f(x)=2 ⋅ (1.34)^x Growth
Step-by-step explanation:
An exponential function at its heart has a base number of rate. If the rate is less than 1, then the function decays. If the base number or rate is greater than 1, then the function grows and increase.
f(x)=(1.023) ⋅ 3^x Rate 3 - Growth
f(x)=3 ⋅ (0.072)^x Rate 0.072 - Decay
f(x)=4 ⋅ (0.035)^x Rate 0.035 - Decay
f(x)=2 ⋅ (1.34)^x Rate 1.34 - Growth
<h2>•5×7^2</h2>
<em>HOPE</em><em> </em><em>ITS</em><em> </em><em>HELPFUL</em><em> </em>^_^
<h2>
•RHONA</h2>
1/2n + 15 = 9 <== ur equation
1/2n = 9 - 15
1/2n = - 6
n = -6 * 2
n = - 12 <== ur solution
