Without using tables, find the value of x in the equation. log x3 + log 5x = 5 log 2 - log (2/5)

15. Without using tables, find the value of x in the equation.
$\log x^3 + \log 5x = 5 \log 2 - \log (2/5)$

• Step 1: Simplify the left-hand side (LHS) of the equation using logarithm properties. The property used is $\log a + \log b = \log (ab)$.

  $$\log x^3 + \log 5x = \log (x^3 \cdot 5x)$$ $$= \log (5x^{3+1})$$ $$= \log (5x^4)$$

• Step 2: Simplify the right-hand side (RHS) of the equation using logarithm properties. The properties used are $n \log a = \log a^n$ and $\log a - \log b = \log (a/b)$.

  $$5 \log 2 - \log (2/5)$$

  First, apply $n \log a = \log a^n$:

  $$= \log 2^5 - \log (2/5)$$ $$= \log 32 - \log (2/5)$$

  Next, apply $\log a - \log b = \log (a/b)$:

  $$= \log \left(\frac{32}{2/5}\right)$$

  To divide by a fraction, multiply by its reciprocal:

  $$= \log \left(32 \times \frac{5}{2}\right)$$ $$= \log (16 \times 5)$$ $$= \log 80$$

• Step 3: Equate the simplified LHS and RHS and solve for x.

  $$\log (5x^4) = \log 80$$

  Since the logarithms are equal and have the same base (assumed to be 10), their arguments must be equal:

  $$5x^4 = 80$$

  Divide both sides by 5:

  $$x^4 = \frac{80}{5}$$ $$x^4 = 16$$

  Take the fourth root of both sides. Since $x$ is part of a logarithm argument ($\log x^3$ and $\log 5x$), $x$ must be positive.

  $$x = \sqrt[4]{16}$$ $$x = 2$$

The final answer is 2.