Ans .
3,-3,-1
(1) let LOG(3, 27)=\(3^3\) or n=3.
ie, \(log_3\)(27)= 3.
(2) Let \(log_7\)(1\343) = n.
Then ,7n =1/343=1/73
n = -3.ie,log7(1\343)= -3.
(3) let \(log_{100}\)(0.01) = n.
Then,. (100) = 0.01 = 1 /100=100 -1 0r n=-1
Ans .
0,0,16Reference https://www.physicsforums.com/threads/how-to-write-log-in-latex.817954/
solution: i) we know that \(log_a\) 1=0 ,so \(log_7\) 1=0 . ii) we know that \(log_a\)a=1,so \(\log_{34} 34\) =0. iii) We know that \(a^{\log_{6} x}\) =x.
now \(36^{\log_{6} 4}\)=\(6^{2^{\log_{6} 4}} \) =\(6^{\log_{6} 16}\)=16.
Ans .
32
\(\log_{\sqrt{8}} x\)=10/3
x=\(\sqrt{8}^{10/3}\) =\(2^{\frac{2}{3} ^{\frac{10}{3}}}\)=\(2^{\frac{2}{3} * \frac{10}{3}}\)=25=32.
Ans .
2/3,5/6
\(\log_{5} 3\)* \(\log_{27} 25 \)=log 3/log 5*log 25/log 27 =(log 3 /log 5) * log\(5^2\) *log\(3^3\)
=(log 3/log 5)*(2log 5 / 3(log 3)
=2/3
(ii)Let \(\log_{9} 27\)=n
Then,\(9 ^n\) =27 \(3^{2n}\) =\(3^3\) 2n=3 n=3/2
Again, let \(\log_{27} 9\)=m
Then,\(27^m\) =9 \(3^{3m}\) =\(3 ^2\) 3m=2 m=2/3
\(\log_{9} 27\)- \(\log_{27} 9\)=(n-m)=(3/2-2/3)=5/6
Ans .
log2
log 75/16-2 log 5/9+log 32/243
= log 75/16-log(5/9)2+log32/243
= log 75/16-log25/81+log 32/243
= log(75/16*32/243*81/25)=log 2
Ans .
7/2
\(\log_{10} 3\)+\(\log_{10} (4x+1)\)=\(\log_{10} (x+1)\)+1
\(\log_{10} 3\)+\(\log_{10} (4x+1)\)=\(\log_{10} (x+1)\)+\(\log_{10} (x+1)\)+\(\log_{10} 10\)
\(\log_{10} (3(4x+1))\)=\(\log_{10} (10(x+1))\) =3(4x+1)=10(x+1)=12x+3 =10x+10 =2x=7=x=7/2
Ans .
2
\(\log_{xyz} (xy)\) + \(\log_{xyz} (yz)\) + \(\log_{xyz} (zx)\)
=\(\log_{xyz} (xy*yz*zx)\)=\(\log_{xyz} {(xyz)^2}\) 2\(\log_{xyz} (xyz)\) =2*1=2
Ans .
1.69897log 50=log (100/2)=log 100-log 2=2-0.30103=1.69897.
Ans .
1.39,0.65
i) log 25=log(100/4)=log 100-log 4=2-2log 2=(2-2*.3010)=1.398.
ii) log 4.5=log(9/2)=log 9-log 2=2log 3-log 2=(2*0.4771-.3010)=0.6532
Ans .
17
log \(2^{56}\) =56 log2=(56*0.30103)=16.85768. Its characteristics is 16. Hence,the number of digits in \(2^{56}\) is 17