Logrithm

Created

Jun 17, 2025

Last Modified

2 months ago

$lo g_{81} 27$
✅ 3/4
$lo g_{36} (\frac{1}{216})$
✅ -3/6
$lo g_{6} (216)$
✅ 6
$lo g_{1/5} (625)$
✅ -4
$lo g_{10} (0.0001)$
✅ -4
$lo g_{0.01} (10000)$
✅ -2
$(lo g_{3} 32) + (lo g_{9} 243)$
✅ 7
$(lo g_{2} 8) + (lo g_{3} 9) + (lo g_{5} 25)$
✅ 8
$lo g_{2} (lo g_{3} 81)$
✅ 2
$lo g_{2} lo g_{2} lo g_{3} lo g_{5} (125)^{3}$
✅ 0
$lo g_{2} lo g_{2} lo g_{2} lo g_{2} (65536)$
✅ 1
$lo g_{5} (lo g_{3} (lo g_{2} x))$ = 1
✅ 2
Question
Hint
$lo g_{10} 80 x$ = 6
✅ 12,500
$lo g_{10} (y) = 6 y = 1 0^{6}$
Question
Explanation
$lo g_{2} lo g_{3} lo g_{2} lo g_{x} 2^{1024} = 1$
✅ 4
$lo g_{3} lo g_{2} lo g_{x} 2^{1024} = 2^{1}$
$lo g_{2} lo g_{x} 2^{1024} = 3^{2}$
$lo g_{x} 2^{1024} = 2^{9}$
$1024 \cdot lo g_{x} 2 = 2^{9}$
$lo g_{x} 2 = 2^{9} / 2^{10}$
$2 = x^{1/2} = x$
$x = 4$
$lo g \frac{14}{13} + lo g \frac{11}{5} - lo g \frac{22}{14}$
✅ 1
$lo g_{e} (x - 1) + l o g_{e} (x) + l o g_{e} (x + 1) = 0$
✅ $x^{3} - x - 1 = 0$
$lo g (\frac{a ^{3}}{b ^{c}}) + lo g (\frac{b ^{3}}{a ^{c}}) + lo g (\frac{c ^{3}}{ab})$
✅ 1
$lo g \frac{15}{8} + 2 lo g \frac{8}{5} - 3 lo g \frac{2}{3} - lo g 2.7$
✅ -1
$lo g_{x} 5 + lo g_{x} 8 + lo g_{x} 12 - lo g_{x} 60$
✅ 3
$lo g_{3} x + lo g_{9} x + lo g_{27} x + lo g_{81} x$
✅ 4
$\frac{1}{2} lo g_{2} P + 2 lo g_{2} Q = 1 + lo g_{0.625} 2$
✅ 4
$lo g_{10} 200 + lo g_{10} 40 + 2 lo g_{10} 25$
✅ 4

Question	Hint
$lo g_{10} 80 x$ = 6 ✅ 12,500	$lo g_{10} (y) = 6 y = 1 0^{6}$

Question	Explanation
$lo g_{2} lo g_{3} lo g_{2} lo g_{x} 2^{1024} = 1$ ✅ 4	$lo g_{3} lo g_{2} lo g_{x} 2^{1024} = 2^{1}$ $lo g_{2} lo g_{x} 2^{1024} = 3^{2}$ $lo g_{x} 2^{1024} = 2^{9}$ $1024 \cdot lo g_{x} 2 = 2^{9}$ $lo g_{x} 2 = 2^{9} / 2^{10}$ $2 = x^{1/2} = x$ $x = 4$

$\frac{1}{lo g _{5} 30} + \frac{1}{lo g _{6} 30}$
$\frac{1}{lo g _{ab} ab c} + \frac{1}{lo g _{b c} ab c} + \frac{1}{lo g _{a c} ab c}$
$If lo g_{10} 2 = 0.3010 then lo g_{2} 10 = ?$
$2 5^{- \frac{1}{4} l o g_{5} 25}$
$4^{l o g_{2} 8} + 2 7^{l o g_{27} 81} = 144 + l o g_{10} x$
$2 lo g_{5} A + \frac{1}{2} lo g_{5} B = 1$
$lo g 3 = 0.4771$ , number of digits in $3^{33}$
$lo g 2 = 0.3010$ , $l o g 3 = 0.4771$ value of $lo g 0.0075 = ?$
$lo g_{2} lo g_{2} lo g_{2} lo g_{3} (6561)$
If $lo g_{y} x = 5 & lo g_{2 y} (8 x) = 4$ then $x = ?$
$1 6^{l o g_{4} 3}$
$\frac{1}{lo g _{3} 2} + \frac{2}{lo g _{9} 4} - \frac{3}{lo g _{27} 8}$
$x^{2} + 7^{l o g_{7} x} - 2 = 0$
Number of digits of $6^{50}$
$4 lo g_{3} 2 - lo g_{3} 4 - 2 lo g_{3} 3 - lo g_{3} 12$
$2 lo g_{10} 20 - (lo g_{10} 5 + lo g_{10} 8)$
$lo g_{3} (3 z + 4) - lo g_{3} z = 2$
$4 lo g_{2} (\frac{1}{4}) - 3 lo g_{\frac{1}{2}} (32)$
$lo g_{3} 4 lo g_{4} 5 lo g_{5} 6 lo g_{6} 7 lo g_{7} 8 lo g_{8} 9$
If a, b, c are consecutive positive integers
$8 lo g (1 + a c) = 2$

value of k = ?