$Log_{b}b$ = 1. $Hence \:log_{5}5$ = 1
$Log_{b}1$ = 0. $Hence\: log_{5}1$ = 0
log(a × b) = log a + log b
similarly, log(a × b × c) = log a + log b + log c
Hence log(2×4×6) = log 2 + log 4 + log 6
log(3+4) = log(3 × 4) is wrong
LHS = log(3+4) = log 7
RHS = log(3 × 4) = log(12)
log 7 ≠ log 12
$log_{5}1024$
$\dfrac{log\:1024}{log\:5}$
=$\dfrac{log\left(2^{10}\right)}{log\left(\dfrac{10}{2}\right)}$
=$\dfrac{10\:log\left(2\right)}{log\:10-log\:2}$
=$\dfrac{10\times0.3010}{1-0.3010}$
=$\dfrac{3.01}{0.699}$
=$\dfrac{3010}{699}$
=4.31
$log(648)^{5}$
= 5 log(648)
= 5 log(81 × 8)
= 5[log(81) + log(8)]
=$5 [log(3^{4}) + log(2^{3})] $
=5[4log(3) + 3log(2)]
= 5[4 × 0.4771 + 3 × 0.30103]
= 5(1.9084 + 0.90309)
= 5 × 2.81149
≈ 14.05
ie, $log(648)^{5}$ ≈ 14.05
ie, its characteristic = 14
Hence, number of digits in $(648)^{5}$ = 14+1 = 15
$log_{4}x+log_{2}x$=12
=>$\dfrac{log\:x}{log\:4}+\dfrac{log\:x}{log\:2}$=12
=>$\dfrac{log\:x}{log\:2^2}+\dfrac{log\:x}{log\:2}$=12
=>$\dfrac{log\:x}{2\:log\:2}+\dfrac{log\:x}{log\:2}$=12
=>$\dfrac{log\:x+2\:log\:x}{2\:log\:2}$=12
=>$\dfrac{3\:log\:x}{2\:log\:2}$=12
Therefore,
logx=$\dfrac{12\times2\:log\:2}{3}$
=8 log 2
=$log(2^8)$
=log(256)
x=256
Let $log_{(.001)} (100)$ =p
$(.001)^p$=100
$\left(\dfrac{1}{1000}\right)^p$=100
$\left(\dfrac{1}{10^3}\right)^p$=$10^2$
$[\left(10\right)^{-3}]^p$=$10^2$
$\left(10\right)^{-3p}$=$10^2$
-3p=2
p=$\dfrac{-2}{3}$
$log_{3} 4\times log_{4} 5 \times log_{5} 6\times log_{6} 7 \times log_{7} 8 \times log_{8} 9 \times log_{9} 9$
$\dfrac{log \:4}{log\:3} \times \dfrac{log \:5}{log\:4} \times \dfrac{log \:6}{log\:5} \times \dfrac{log \:7}{log\:6} \times \dfrac{log \:8}{log\:7} \times \dfrac{log \:9}{log\:8} \times 1$
=$\dfrac{log \:9}{log\:3}$
=$\dfrac{log \:3^2}{log\:3}$
=$\dfrac{2log \:3}{log\:3}$
=2
$log_{12} 27$ = a
=> log 27/ log 12 = a
=> $log 3^{3}$ / $log (3 * 2^{2})$ =a
=> 3 log 3 / log 3 + 2 log 2 = a
=> (log 3 + 2 log 2)/ 3 log 3 = 1/a
=> (log 3/ 3 log 3) + (2 log 2/ 3 log 3) = 1/3
=> (2 log 2)/ (3 log 3) = 1/a – 1/3 = (3-a)/ 3a
=> log 2/ log 3= (3-a)/3a
=> log 3 = (2a/3-a)log2
$log_{16} 16 $= log 16/ log 6
= $log 2^{4}/ log (2*3) $
= 4 log 2/ (log 2 + log 3)
= 4(3-a)/ (3+a)
$log_{a} (ab)$ = x
=> log b/ log a = x
=> (log a + log b)/ log a = x
1+ (log b/ log a) = x
=> log b/ log a = x-1
log a/ log b = 1/ (x-1)
=> 1+ (log a/ log b) = 1 + 1/ (x-1)
(log b/ log b) + (log a/ log b) = x/ (x-1) => (log b + log a)/ log b = x/ (x-1)
=>log (ab)/ log b = x/(x-1) => $log_{b} (ab)$ = x/(x-1)
= $\dfrac{1}{3}log_{10}125−2log_{10}4+log_{10}32$
=$log_{10}(125^{1/3})−log_{10}(4^2)+log_{10}32$
=$log_{10}5−log_{10}16+log_{10}32$
=$log_{10}\left(\dfrac{5\times32}{16}\right)$
=$log_{10}10$
=1
If base is not mentioned, then always remember to take it as 10.
Hence, in the given expression, assume base as 10
We are given, $[log_{10} 2 + log_{10} (4x + 1)$ = $log_{10} (x + 2) + 1]$
$[log_{10} 2 + log_{10} (4x + 1)$ = $log_{10} (x + 2) + 1]$
$[log_{10} 2 + log_{10} (4x + 1)$ = $log_{10} (x + 2) + log_{10}10]$
Now, Use the product rule: $log_{a}(xy)$ = $log_{a}x + log_{a}y$
$[log_{10} 2 (4x + 1)]$ = $[log_{10} 10(x + 2)]$
(4x + 1) = (5x + 10)
4x + 1 = 5x + 10
x = - 9