The incorrect step is ln(x²) = ln(3x/0) because ln(x/y) = lnx - lny and the solutions are x = 0, 3
<h3>What is a logarithm? </h3>
It is another way to represent the power of numbers, and we say that 'b' is the logarithm of 'c' with base 'a' if and only if 'a' to the power 'b' equals 'c'.
The question is incomplete.
The complete question is in the picture, please refer to the attached picture.
We have:
2ln(x) = In(3x) - [In(9) - 2ln(3)]
In(x²) = In(3x) - [In(9) - In(9)]
ln(x²) = ln(3x) - 0
ln(x²) = ln(3x/0) (incorrect step)
ln(x²) = ln(3x)
x² = 3x (correct step)
x² - 3x =0
x(x - 3) = 0
x = 0, 3
Thus, the incorrect step is ln(x²) = ln(3x/0) because ln(x/y) = lnx - lny and the solutions are x = 0, 3
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Answer:
Step-by-step explanation:
The 9th term is the next-to-last term, where the left term of the binomial is raised to the first power, the right term of the binomial is raised to the 8th power (9-1=8), and the multiplier is 9C8 = 9!/(8!·1!) = 9. This product is ...
9·(3x)^1·(-2y)^8 = 6912xy^8
Answer:
£43
Step-by-step explanation:
925×2/100
=£ 18.5
2.4 years =18.5×2.4
=£43
<h2>
Answer:</h2>
<h2>
Step-by-step explanation:</h2>
a. 2x^-3 • 4x^2
To solve this using only positive exponents, follow these steps:
i. Rewrite the expression in a clearer form
2x⁻³ . 4x²
ii. The position of the term with negative exponent is changed from denominator to numerator or numerator to denominator depending on its initial position. If it is at the numerator, it is moved to the denominator. If otherwise it is at the denominator, it is moved to the numerator. When this is done, the negative exponent is changed to positive.
In our case, the first term has a negative exponent and it is at the numerator. We therefore move it to the denominator and change the negative exponent to positive as follows;
iii. We then solve the result as follows;
=
Therefore, 2x⁻³ . 4x² =
b. 2x^4 • 4x^-3
i. Rewrite as follows;
2x⁴ . 4x⁻³
ii. The second term has a negative exponent, therefore swap its position and change the negative exponent to a positive one.
iii. Now solve by cancelling out common terms in the numerator and denominator. So we have;
Therefore, 2x⁴ . 4x⁻³ =
c. 2x^3y^-3 • 2x
i. Rewrite as follows;
2x³y⁻³ . 2x
ii. Change position of terms with negative exponents;
iii. Now solve;
Therefore, 2x³y⁻³ . 2x =
f(x) - g(x)
=> (3x² + x - 3) - (x² - 5x + 1)
=> 3x² + x - 3 - x² + 5x - 1
=> 3x² - x² + x + 5x - 3 - 1
=> 2x² + 6x - 4
