Answer:
1. The given equation is presented as follows;
[tex]3^{2 \cdot x - 1} = \dfrac{1}{9}[/tex]
The above equation can be rewritten as follows;
[tex]3^{2 \cdot x - 1} = 3^{-2}[/tex]...(1)
Therefore, taking logarithm to base 3 of both sides of the equation, the above equation can be rewritten without an exponent as follows;
[tex]log_3 \left (3^{2 \cdot x - 1} \right) =log_3\left(\dfrac{1}{9} \right) = log_3 \left( 3^{-2} \right)[/tex]...(2)
2. Inverse relationship of exponents is given as follows;
(1/a)ˣ = a⁻ˣ
Therefore, we have;
1/9 = 1/(3²) = (3²)⁻¹ = 3⁻²
Taking log of both sides of [tex]\left (3^{2 \cdot x - 1} = \dfrac{1}{9} = 3 ^{-2}\right)[/tex] gives;
[tex]log_3 \left (3^{2 \cdot x - 1} \right)= log_3 \left( 3^{-2} \right)[/tex]
From logarithm rules, we have;
From equation (2), we have;
[tex](2 \cdot x - 1) \cdot log_3 3 = -2 \cdot log_3 3[/tex]
Therefore, the equation can be simplified by dividing both sides by the common factor, [tex]log_3 3[/tex], as follows;
[tex]\dfrac{(2 \cdot x - 1) \cdot log_3 3 }{log_3 3} = \dfrac{-2 \cdot log_3 3}{log_3 3}[/tex]
Therefore, we get;
2·x - 1 = -2...(3)
3. Solving 2·x - 1 = -2, gives;
2·x - 1 = -2
2·x = -2 + 1 = -1
x = -1/2 = -0.5
x = -0.5
EQUATION 3
The given equation is presented as follows;
[tex]125^{x - 1} = 5^x[/tex]
1. By the property of exponents, we have;
[tex]\left(a^b \right)^x = a^{b \times x}[/tex]
125 = 5³
Therefore, we get;
[tex]125^{x - 1}= \left(5^3\right)^{x - 1} = 5^{3\times (x - 1)} = 5^{(3\times x - 3)}[/tex]
[tex]125^{x - 1}= 5^{(3\times x - 3)}[/tex]
2. Given that we can rewrite the given equation as follows;
[tex]125^{x - 1} = 5^{(3\times x - 3)} = 5^x[/tex]
Therefore;
[tex]5^{(3\times x - 3)} = 5^x[/tex]
Both sides of the equation are given in powers of 5, therefore, the appropriate base logarithm to use on both sides to rewrite the equation is logarithm to base 5
3. Using logarithm, by applying logarithm to both sides, we get;
[tex]log_5 \left (5^{(3\times x - 3)} \right) = log_5 \left(5^x \right)[/tex]
[tex](3\times x - 3) \cdot log_5 5 = x \cdot log_5 5[/tex]
Dividing both sides by log₅5 gives;
[tex]\dfrac{ (3\times x - 3) \cdot log_5 5}{log_5 5} = \dfrac{ x \cdot log_5 5}{log_5 5}[/tex]
∴ 3 × x - 3 = x
3 × x - x = 3
2·x = 3
x = 3/2 = 1.5
x = 1.5
Step-by-step explanation:
