Name: Logarithmic Inequalities
Uploaded: 2019-07-19
Duration: 10 min 24 s
Description: Logarithmic inequalities are inequalities in which one (or both) sides involve a logarithm. Watch this video and learn how to solve logarithmic inequalities.

Question

If

\log_{12} 18 = α

and

\log_{24} 54 = β

. Prove that :

α β + 5 (α - β) = 1

.

Accepted Answer

Given that $\log_{12} 18 = α . . . (1)$

and $\log_{24} 54 = β . . . (2)$

We have to prove that $α β + 5 (α - β) = 1 . . . (3)$

Firstly, we will use the change of base rule $\log_{a} b = \frac{\log b}{\log a}$ on both equations (1) and (2), we get

$\frac{\log 18}{\log 12} = α$

$\Rightarrow \frac{\log (3^{2} \times 2)}{\log (2^{2} \times 3)} = α . . . (4)$

and $\frac{\log 54}{\log 24} = β$

$\Rightarrow \frac{\log (3^{3} \times 2)}{\log (2^{3} \times 3)} = β . . . (5)$

Now, we will apply the product rule $\log (m n) = \log m + \log n$ and power rule $\log (m^{n}) = n \log m$ on equations (4) and (5), we get

$α = \frac{2 \log 3 + \log 2}{2 \log 2 + \log 3} . . . (6)$

and $β = \frac{3 \log 3 + \log 2}{3 \log 2 + \log 3} . . . (7)$

Let us take $\log 2 = x$ and $\log 3 = y$ and substitute these values in equations (6) and (7), we get

$α = \frac{2 y + x}{2 x + y}$ and $β = \frac{3 y + x}{3 x + y}$

Further, we will substitute these values on the left-hand side of the equation (3), we get

$α β + 5 (α - β) = (\frac{2 y + x}{2 x + y}) (\frac{3 y + x}{3 x + y}) + 5 ((\frac{2 y + x}{2 x + y}) - (\frac{3 y + x}{3 x + y}))$

$\Rightarrow α β + 5 (α - β) = \frac{6 y^{2} + 5 x y + x^{2}}{6 x^{2} + 5 x y + y^{2}} + 5 (\frac{3 x^{2} + 7 x y + 2 y^{2} - (2 x^{2} + 7 x y + 3 y^{2})}{6 x^{2} + 5 x y + y^{2}})$

$\Rightarrow α β + 5 (α - β) = \frac{6 y^{2} + 5 x y + x^{2}}{6 x^{2} + 5 x y + y^{2}} + \frac{5 x^{2} - 5 y^{2}}{6 x^{2} + 5 x y + y^{2}}$

$\Rightarrow α β + 5 (α - β) = \frac{6 x^{2} + 5 x y + y^{2}}{6 x^{2} + 5 x y + y^{2}}$

Thus, it is proved that $α β + 5 (α - β) = 1$

If $\log_{12} 18 = α$ and $\log_{24} 54 = β$ . Prove that : $α β + 5 (α - β) = 1$ .

Important Questions on Appendix II : Logarithm

Learn and Prepare for any exam you want !

If log1218=α and log2454=β. Prove that : αβ+5α-β=1.

Important Questions on Appendix II : Logarithm

Learn and Prepare for any exam you want !

If $\log_{12} 18 = α$ and $\log_{24} 54 = β$ . Prove that : $α β + 5 (α - β) = 1$ .