Question

X! Log7base14,Y=log14base21,Z=log21base28then 1+XyZ

Answer 1

please see the attachment

Answer 2

The value of 1+XYZ is $\log_{28} 196$

Given : X = $\log_{14}7$  , Y  = $\log_{21}14$ , Z = $\log_{28}21$

To Find : The value of 1+XYZ

Solution : The value of 1+XYZ is $\log_{28} 196$

We know the property of logarithm function is

$\log_{a}b = \frac{\log_e b}{\log_e a}$

Now using the property in the above question

we have to find the find the value of 1+XYZ

where X = $\log_{14}7$  

Y  = $\log_{21}14$

Z = $\log_{28}21$

putting the values of X , Y and Z in the above equation

1 + ( $\log_{14}7$ ) ( $\log_{21}14$ ) ( $\log_{28}21$)

using the above property

we have

1 + ($\frac{\log_e 7}{\log_e 14}$ ×  $\frac{\log_e 14}{\log_e 21}$   × $\frac{\log_e 21}{\log_e 28}$)

= 1 + ($\frac{\log_e 7}{\log_e 28}$)

= $\frac{\log_e 28 + \log_e 7}{\log_e 28}$   (by using the property that $\log a+\log b = \log ab$)

= $\frac{\log_e 196}{\log_e 28}$       ( by using the property that $\log_{a}b = \frac{\log_e b}{\log_e a}$)

= $\log_{28} 196$

The value of 1+XYZ is $\log_{28} 196$

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