Question

LOGARITHMIC Equation:

Answer 1

used identity are
log mⁿ = n log m

log base m (n) = 1/log base n (m)
log m + log n = log mn

thus

(2/log base 4 (2000)^6) + (3/log base 5 (2000)^6 )


(2/6)log base 2000 (4) + (3/6) *log base 2000 (5)

=( 2log base 2000 (4) +3 log base 2000 (5) )/6

=( log base 2000 (16) + log base 2000 (125))/6

=( log base 2000. (16*125))/6

= (log base 2000 (2000))/6

= 1/6
thus answer will be 1/6

Answer 2

Hey !!!

$\frac{2}{ log_{4}(2000) ^{6} } + \: \frac{3}{ log_{5}(2000) ^{6} }$


Using rule :- change of base
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$2 log_{2000 ^{6} } (4) + \: 3 log_{2000 ^{6} }(5)$



$\frac{1}{6} log_{2000}(4 ^{2} ) + \: \frac{1}{6} log_{2000}(5 )^{3} )$


Base are same here

hence ,

$\frac{1}{6} log_{2000}(5 ^{3} \times 4 ^{2} )$

$\frac{1}{6} log_{2000}(2000)$


1 / 6 × 1

1 / 6 Answer ✔

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Hope it does't help you !!!

@Rajukumar111