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Evaluate 2 128, simplify 3 ((x2)/(sqrt(z))), and solve for x: 52x−1 = 125.

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3 parts
a

7\mathbf{7}

b

2log⁡3x−12log⁡3z\mathbf{2\log_3 x - \frac{1}{2}\log_3 z}

c

2\mathbf{2}

Step-by-step solution

10 steps

Here's the solution for question 4:

Part a· 3 steps

Evaluate log⁡2128\log_2 128.

  1. Express 128 as a power of 2.

    128=2×2×2×2×2×2×2=27128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7

  2. Substitute this into the logarithm expression.

    log⁡2128=log⁡2(27)\log_2 128 = \log_2 (2^7)

  3. Use the logarithm property log⁡b(bk)=k\log_b (b^k) = k.

    log⁡2(27)=7\log_2 (2^7) = 7

    The final answer is 7\boxed{\mathbf{7}}.

Part b

2log⁡3x−12log⁡3z\mathbf{2\log_3 x - \frac{1}{2}\log_3 z}

Simplify log⁡3(x2z)\log_3 \left(\frac{x^2}{\sqrt{z}}\right).

  1. Apply the logarithm property log⁡b(MN)=log⁡bM−log⁡bN\log_b \left(\frac{M}{N}\right) = \log_b M - \log_b N.

    log⁡3(x2z)=log⁡3(x2)−log⁡3(z)\log_3 \left(\frac{x^2}{\sqrt{z}}\right) = \log_3 (x^2) - \log_3 (\sqrt{z})

  2. Rewrite z\sqrt{z} as z1/2z^{1/2}.

    =log⁡3(x2)−log⁡3(z1/2)= \log_3 (x^2) - \log_3 (z^{1/2})

  3. Apply the logarithm property log⁡b(Mk)=klog⁡bM\log_b (M^k) = k \log_b M.

    =2log⁡3x−12log⁡3z= 2\log_3 x - \frac{1}{2}\log_3 z

    The final answer is 2log⁡3x−12log⁡3z\boxed{\mathbf{2\log_3 x - \frac{1}{2}\log_3 z}}.

Part c

2\mathbf{2}

Solve for xx: 52x−1=1255^{2x-1} = 125.

  1. Express 125 as a power of 5.

    125=5×5×5=53125 = 5 \times 5 \times 5 = 5^3

  2. Substitute this into the equation.

    52x−1=535^{2x-1} = 5^3

  3. Since the bases are equal, equate the exponents.

    2x−1=32x-1 = 3

  4. Solve the linear equation for xx.

    2x=3+12x = 3+1 2x=42x = 4 x=42x = \frac{4}{2} x=2x = 2

    The final answer is 2\boxed{\mathbf{2}}.

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