If log 18 to the base 12=a and log 54 to the base 24=b,prove that ab+5(a-b)=1?

Jul 11, 2017

see explanation

Explanation:

Please check to make sure I've read the problem correctly. Logarithms may be written wrong...

1. $\setminus {\log}_{12} \left(18\right) = a$
2. $\setminus {\log}_{24} \left(54\right) = b$

ab+5(a-b)\stackrel{?}{=}1
$\left(\setminus {\log}_{12} \left(18\right) \setminus \cdot \setminus {\log}_{24} \left(54\right)\right) + 5 \left(\setminus {\log}_{12} \left(18\right) - \setminus {\log}_{24} \left(54\right)\right) \setminus \stackrel{\setminus \textcolor{o l i v e}{\setminus \sqrt{}}}{=} 1$

Jul 12, 2017

Explanation:

As ${\log}_{12} \left(18\right) = a$, we have ${12}^{a} = 18$

i.e. ${\left({2}^{2} \times 3\right)}^{a} = 2 \times {3}^{2}$

or ${2}^{2 a - 1} \times {3}^{a - 2} = 1$

similarly as ${\log}_{24} \left(54\right) = b$, we have ${24}^{b} = 54$

i.e. ${\left({2}^{3} \times 3\right)}^{b} = 2 \times {3}^{3}$

or ${2}^{3 b - 1} \times {3}^{b - 3} = 1$

Comparing the two $2 a - 1 = 3 b - 1$ or $2 a - 3 b = 0$ ........(A)

and $a - 2 = b - 3$ or $a - b + 1 = 0$ i.e. $2 a - 2 b + 2 = 0$ ........(B)

Subtracting (A) from (B), we get

$b + 2 = 0$ i.e. $b = - 2$

and $a = b - 1 = - 3$

hence $a b + 5 \left(a - b\right)$

$= \left(- 3\right) \times \left(- 2\right) + 5 \times \left(- 1\right) = 6 - 5 = 1$

(Note that (B) gives $a - b = - 1$.)