Question

Show that:
$\log_{y} \sqrt{x} \cdot \log_{z} y^{3} \cdot \log_{x} \sqrt[3]{z^{2}} = 1$

Answer 1

Answer:

Step-by-step explanation:

Hi,

Here we will be using the following properties of

logarithm:

logₐb = log b/log a

nlog a = log aⁿ

Consider L.H.S

= log y(√x) . logz(y³).logₓ∛z²

log y(√x) = log √x/log y

But log √x = 1/2 log x,

So log y(√x) =  1/2* log x/log y

logz(y³) = log y³/log z

log y³ = 3 log y

So, logz(y³) = 3 log y/log z

logₓ∛z² = log  ∛z²/log x = log z²⁾³ = 2/3 log z

So, logₓ∛z² = 2/3 log z/log x

Substituting the above in L.H.S, we get

L.H.S = (1/2* log x/log y)*(3 log y/log z)*(2/3 log z/log x)

= 1/2*3*2/3*

= 1

= R.H.S

Hence, Proved

Hope, it helps !

Answer 2

Solution :

$\textsf { We know the Logarithmic laws }$
$\log_{b}a = \frac{loga}{logb}$
$\log_{a}n^{m} = mlog_{a}n$
$\ log_{a}a = 1$

LHS =

$\log_{y} \sqrt{x} \cdot \log_{z} y^{3} \cdot \log_{x} \sqrt[3]{z^{2}}$

= $\left ( \frac{1}{2} \right )\cdot \log_{y}x \cdot \left ( \frac{3}{1} \rifht )\cdot \log_{z}y \cdot \left (\frac{2}{3} \right ) \log_{x}{z}}$

= $\Big (\frac{1.3.2}{2.1.3} \Big ) \cdot \Big (\frac{logx}{logy} \cdot \frac{logy}{logz}\cdot \frac{logx}{logz} \Big )$

= $ 1×1 $

= $1$

= $\textsf { R.H.S }$

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