Answer:
its actually 16
Step-by-step explanation:
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:
Jorge was charged for 10 hours that month
Step-by-step explanation:
Given : Jorge’s monthly bill from his Internet service provider was $25.
Also given : The service provider charges a base rate of $15 per month plus $1 for each hour that the service is used
We have to find the number of hours that Jorge was charged for that month.
Let the number of hour be x,
Total bill = base rate + 1 (number of hours used)
Total bill = base rate + 1(x)
Substitute the values, we get,
⇒ 25 = 15 + 1 (x)
Solving we get,
⇒ 25 - 15 = x
10 = x
Thus, Jorge was charged for 10 hours that month
