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.
It’s cheaper to take this bus. Compare 166 to 190
I think it’s always because the absolute value is always positive.
Answer:
15 kg
Step-by-step explanation:
Radius of a solid sphere : mass of a solid sphere
3 cm : 9 kg
Find the mass of a sphere of the same material of radius 5 cm
Radius of a solid sphere : mass of a solid sphere
5 cm : x kg
Equate both ratios to find x
3 cm : 9 kg = 5 cm : x kg
3/9 = 5/x
3 * x = 9 * 5
3x = 45
x = 45/3
x = 15 kg
The mass of a sphere of the same material of radius 5 cm is 15 kg
