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Indices and Logarithms

Evaluate:

\( 27^{\frac{2}{3}} \times \left( \frac{81}{16} \right)^{-\frac{1}{4}} \)

Solution:

\( \left( \sqrt[3]{27} \right)^2 \times \left( \frac{16}{81} \right)^{\frac{1}{4}} \)

\( (3)^2 \times \frac{2}{3} \)

\( 3^2 \times \frac{2}{3} = \underline{6} \).

Solve the equation:

\( 9^{t+1} + 3^{2t} = 30 \)

Solution:

\( 3^{2(t+1)} + 3^{2t} = 30 \)

\( 3^{2t+2} + 3^{2t} = 30 \)

\( 3^{2t} \cdot 3^2 + 3^{2t} = 30 \)

\( 3^{2t} (3^2 + 1) = 30 \)

\( 3^{2t} \cdot 10 = 30 \)

\( 3^{2t} = 3^1 \)

\( 2t = 1 \)

\( t = \frac{1}{2} \)

16. Solve for \( k \) in the following equation:

\( 125^{k+1} + 5^{3k} = 630 \)

Solution:

\( 5^{3(k+1)} + 5^{3k} = 630 \)

\( 5^{(3k+3)} + 5^{3k} = 630 \)

Using the exponent rule: \( a^m \times a^n = a^{(m+n)} \), we rewrite:

\( 5^{3k} \cdot 5^3 + 5^{3k} = 630 \)

\( 5^3 \cdot 5^{3k} + 5^{3k} = 630 \)

\( 125 (5^{3k}) + 5^{3k} = 630 \)

Let \( 5^{3k} = t \), then:

\( 125t + t = 630 \)

\( 126t = 630 \)

\( t = 5 \)

\( 5^{3k} = 5^1 \)

\( 3k = 1 \)

\( k = \frac{1}{3} \)

Solve for \( x \) in the equation (3 marks):

\( 5^{2x-1} - 25^x = 500 \)

Solution:

\( 5^{2x} \cdot 5^{-1} - 5^{2x} = 500 \)

Let \( 5^{2x} = y \), then:

\( y \cdot \frac{1}{5} - y = 500 \)

\( \frac{1}{5}y - y = 500 \)

Factor out \( y \):

\( y \left( \frac{1}{5} - 1 \right) = 500 \)

\( y \left( -\frac{4}{5} \right) = 500 \)

Multiply both sides by \( -\frac{5}{4} \):

\( -4y = 2500 \)

\( y = 625 \)

\( 5^{2x} = -625 \)

\( 5^{2x} = -5^4 \) but a power of 5 cannot be negative,

therefore there is no real solution for \( x \).

Solve for the value of \( x \)

\( 2^{3x-2} \times 8^x = 4^{x+1} \)

\( 2^{3x-2} \times 2^{3x} = 2^{2(x+1)} \)

\( (3x - 2) + 3x = 2x + 2 \)

\( 6x - 2x = 2 + 2 \)

\( 4x = 4 \)

\( x = 1 \)

Given that \( \sqrt[3]{9^4} = 3^n \), find the value of \( n \).

Solution:

\(\sqrt[3]{a} = a^{\frac{1}{3}}\)

\(\sqrt[3]{9^4} = 9^{\frac{4}{3}}\)

Since \( 9 = 3^2 \), we rewrite:

\( 9^{\frac{4}{3}} = (3^2)^{\frac{4}{3}} \)

Using exponent rules: \( (a^m)^n = a^{m \cdot n} \)

\( 3^{2 \times \frac{4}{3}} = 3^{\frac{8}{3}} \)

Thus, comparing exponents:

\( 3^{\frac{8}{3}} = 3^n \Rightarrow n = \frac{8}{3} \)

\( n = 2 \frac{2}{3} \)

\( 64^{\frac{3}{2}} \)

\( = (64^{\frac{1}{2}})^3 \)

\( = (\sqrt{64})^3 \)

\( = 8^3 \)

\( = 512 \)

\(\text{Simplify without using a calculator}\)

\( 4^{\frac{5}{2}} + 64^{\frac{2}{3}} - 2^0 \)

\( \left(\sqrt{4}\right)^5 + \left(\sqrt[3]{64}\right)^2 - 1 \)

\( 2^5 + 4^2 - 1 \)

\( 32 + 16 - 1 \)

\( 32 + 15 = 47 \)

\( \frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}} \)

\( \frac{2^{3 \times \frac{1}{3}} \times 5^{\frac{2}{3}}}{(2 \times 5)^{\frac{2}{3}}} \)

\( \frac{2^1 \times 5^{\frac{2}{3}}}{2^{\frac{2}{3}} \times 5^{\frac{2}{3}}} \)

\( 2^{1 - \frac{2}{3}} \times 5^{\frac{2}{3} - \frac{2}{3}} \)

\( 2^{\frac{1}{3}} \times 5^0 \)

\( 2^{\frac{1}{3}} = \sqrt[3]{2} \)

2. Solve the equation: \( 27^x + 3^{3x-1} - 4 = 104 \) (3 marks)

\( 27^x + 3^{3x-1} = 104 + 4 = 108 \)

Since \( 27 = 3^3 \),

\( 3^{3x} + \frac{3^{3x}}{3} = 108 \)

Let \( y = 3^{3x} \),

\( y + \frac{y}{3} = 108 \)

Multiply through by 3,

\( 3y + y = 324 \)

\( 4y = 324 \) so \( y = 81 \)

Substituting back,

\( 3^{3x} = 81 \) so \( 3^{3x} = 3^4 \)

Equating the indices,

\( 3x = 4 \) so \( x = \frac{4}{3} = 1\frac{1}{3} \).