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:
93.53m.
Step-by-step explanation:
.
Answer:
Step-by-step explanation:
when y=4
x=4-3=1
when y=8
x=8-3=5
when y=5
x=5-3=2
Answer:
i think it is 1.5
Step-by-step explanation
because 5 tenths is equal .5 and adding 1 unit which is adding on one which should equal 1.5
Answer:
$4
Step-by-step explanation:
Discount is a form of consideration in price or amount given to a customer on a sale of goods. It can be deduced from the question that;
The original price of each plant = $x
Number of plants purchased = 20
Price of the number of plants purchased without discount = $20x
Amount paid for the 20 plants with discount = $(20x - 80)
Thus,
The discount on the 20 plants = $80
Discount on each plant = 
= 
= $4
Discount on each plant = $4
Therefore, each plant was discounted by $4.
