Given that $f(x) = 6x - 5$ and $g(x) = \frac{x + 5}{6}$, we need to investigate if $f(g(x)) = g(f(x))$ - Leaving Cert Mathematics - Question 3 - 2020

To convert, we can take logarithms:

• First, take $\ ext{log}_5$ of both sides: $\log_5 y = \log_5(5x^2)$
• Using log properties: $\log_5 y = \log_5 5 + \log_5(x^2)$
• This simplifies to: $\log_5 y = 1 + 2 \log_5 x$
• Comparing with $\log_5 y = a + b \log_5 x$, we find that $a = 1$ and $b = 2$.

Values:

• $a = 1$
• $b = 2$

• Start by applying logarithm laws: $\log_5 y = \log_5(5^2) + \log_5\left(\frac{126}{25}x - 1\right)$
  • This can be simplified to: $\log_5 y = \log_5\left(25\left(\frac{126}{25}x - 1\right)\right)$
• Hence, setting the arguments equal gives: $y = 25\left(\frac{126}{25}x - 1\right)$
  • Further simplifying: $y = 126x - 25$
  • To find values of $y$: $y = 5x^2 = 126x - 25$
  • Rearranging gives: $5x^2 - 126x + 25 = 0$
• Now, applying the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
  • Where:
  • $a = 5$, $b = -126$, $c = 25$ gives: $x = \frac{126 \pm \sqrt{(-126)^2 - 4 \cdot 5 \cdot 25}}{2 \cdot 5}$
  • Thus solving the quadratic equation provides the values for $x$, and substituting back calculates $y$.