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<u>Answer:</u>
The correct answer option is: True.
<u>Step-by-step explanation:</u>
Its true that if there is no linear equation to start with, you can isolate and substitute a variable that is squared in both the equation.
For example, for the given non linear equation, start by dividing both sides by coefficient of the variable.
Once you do that and isolate a variable, continue solving by substituting that variable into the other equation.
Answer:
9984000
Step-by-step explanation:
subtract the total by the other number (coverage)
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.
1.8 = 18/10
Then the number that’s equal to the 3/5 part of 1.8 is :
(18/10) multiplied by (3/5)
(18/10)(3/5) = 54/50 = 108/100 = 1.08
Then our number is 1.08