Question

Log√27- log√8/log3- log 2

Answer 1

Step-by-step explanation:

Consider the LHS as:

:\frac{\log {\sqrt27} + \log 8 + :\log \sqrt{1000}}{\log{120}}

= :\frac{\log {27^\frac{1}{2}} + \log 2^3 + \log {1000^\frac{1}{2}}}{\log{(2 \times 2 \times 2 \times 3 \times 5)}}

Using the laws of logarithmic which states, \log m^n = n \log m

= \frac{\frac{1}{2} \log(3^3) + \log(2^3) + \frac{1}{2} \log(10^3)}{\log(2^3 \times 3 \times 5)}

=\frac{\frac{3}{2} \log3 + 3 \log(2) + \frac{3}{2} \log(10)}{\log(2^3)+\log3+ \log 5}

= \frac{\frac{3}{2} \log3 + 3 \log(2) + \frac{3}{2} \log(2 \times 5)}{3 \log2+\log3+ \log 5}

= \frac{\frac{3}{2} \log3 + 3 \log2 + \frac{3}{2} (\log2 + \log 5)}{3 \log2+\log3+ \log 5}

= \frac{\frac{3}{2} \log3 + 3 \log2 + \frac{3}{2} \log2 +\frac{3}{2} \log 5}{3 \log2+\log3+ \log 5}

= \frac{\frac{3}{2} \log3 + \frac{9}{2} \log2 +\frac{3}{2} \log 5}{3 \log2+\log3+ \log 5}

= \frac{3}{2}\frac{( \log3 + 3 \log2 + \log 5)}{3 \log2+\log3+ \log 5}

= \frac{3}{2}

=RHS

Hence, proved.

Answer 2

Given :       log√27- log√8/log3- log 2

To find :  Value

Solution:

(log√27- log√8 )/ (log3- log 2)

√27  = √3³   =  $(3^3)^{\frac{1}{2}} = 3^{\frac{3}{2}}$

√8  = √2³   =  $(2^3)^\frac{1}{2} = 2^\frac{3}{2}$

now using  log aᵇ  = b log a  

log √27   = (3/2) log 3

log √8  =  (3/2) log 2

(log√27- log√8 )/ (log3- log 2)

= ((3/2) log 3 -  (3/2) log 2 ) /   (log3- log 2)

= (3/2) (log3- log 2) /  (log3- log 2)

= 3/2

log√27- log√8/log3- log 2 = 3/2

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