Hello, I got stuck with these simple log problems. Can someone solve these problems with detail explanation. 1. log5 10 X log10 15 X log15 20 X log20 25 Ans is 2 2. If 2logx (x-2)=logx 4, find the value of x. Ans is 4 3. Solve the following equations. (i) log 5/4 + log 14 - log 7X/3 = -1 Ans is 75 4. If log10 (x-2)^2 = 4 then value of x is Ans is 102 5. If log10 X = 5 then number of digits in number X are __________________
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2. If 2logx (x-2)=logx 4, find the value of x.
Solition: logx(x-2)^2=logx4
(x-2)^2=4
(x-2)=2
x=4
Ans: 4
"To get the BEST, be the BEST"
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4. If log10 (x-2)^2 = 4 then value of x is
Sol: log 10 (x-2)^2 = 4
log 10 (x-2)^2= 4 log 10 10
= log 10 10^4
So,(x-2) ^2=10^4
(x-2) =100
x= 102
Ans: 102
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Soln.:
log5/log10 * log10/log15 * log15/log20 * log20/log25[Formula logab=logb/loga
=log5/log25
=log5/2log5
=1/2 [Not 2 I think]
Giv ur best to d world.
Nd d best will cm back to u....
Regards,
Dipanjan.....
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5. If log10 X = 5 then number of digits in number X are-
Ans: 6
"To get the BEST, be the BEST"
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Soln.:
log7x/3=log5/4 + log14 + log1010 [ As log1010=1]
log7x/3=log(5/4*14*10)[Formula logm + logn = log(mn)]
=>x=5/4*14*10*3/7=75.
Giv ur best to d world.
Nd d best will cm back to u....
Regards,
Dipanjan.....
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3. Solve the following equations.
(i) log 5/4 + log 14 - log 7X/3 = -1
Sol: log5- log4+ log 14-log 7x+log3+log10=0
[(5*14*3*10)/(4*7x)] =1
Solving,x=75
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Soln.
1/2log(2x+2) +1/2log(3x+4)=log 10 + log2 [ as logx½=1/2logx]
=>log[(3x+2)(2x+4)]^1/2=log20
=>(3x+2)(2x+4)=400
Solve and get x=7...
Giv ur best to d world.
Nd d best will cm back to u....
Regards,
Dipanjan......
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1. log5 10 X log10 15 X log15 20 X log20 25
ans:-
=(log5\log10)*(log10\log15)*(log15\log20)log20\log25)
(b'coz loga b=loga\log b vid base 10)
=log25\log5 (res tes vil be cut out)
=2log5\log5(b'coz loga^b=bloga base is 10 ven not mentioned)
=1
2. If 2logx (x-2)=logx 4, find the value of x.
ans:-
=>logx (x-2)^2=logx 4(by above mentiones property)
=>log((x-2)^2)\logx=log4\logx(prop metioned in prv. queston)
=>log(x-2)^2=log4
=>(x-2)^2=4
taking square root on both sides we get
=>x-2=2
=>x=4
3. Solve the following equations.
(i) log 5/4 + log 14 - log 7X/3 = -1
(ii) 1/2 log (2X+2) + log root of (3X + 4) = 1 + log 2
ans:-
i) =>log((5/4)*(14)/(7x/3))= -1*log10 (b'oz loga+logb-logc=log(a*b/c)
=>15/2x=1/10
=>x=75
ii) =>log(2x+2)^1/2+log(3x+4)^1/2=log10+log2
=>log((2x+2)(3x+4))^1/2=log20
=>3x^6+7x+4=200
=>x=7
4. If log10 (x-2)^2 = 4 then value of x is
=>log10 (x-2)^2 = 4*log10 10
=>(x-2)^2=10^4
taking squre root both sides we get
=>x-2=100
=>x=102
5. If log10 X = 5 then number of digits in number X are
=>log10 x=5*log10 10
=>x=10^5=100000
hence no. of digt are 6
ii) 1/2 log (2X+2) + log root of (3X + 4) = 1 + log 2
Soln:log(2x+2)^ 1/2 +log (3x+4)^1/2=log 10+log 2
log[(2x+2)^1/2 * (3x+4)^1/2] =log(20)
(2x+2)^1/2 *(3x+4)^1/2 =20
(2x+2)(3x+4)=400
6x^2+14x-392=0
x=-56/6 and 7.
x cannot be -56/6 so 7.
Ans:7
"To get the BEST, be the BEST"
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Hello,
Thanks Shivani for the reply.
Can you pls be elaborate on how you got the answer as 6 for Q5. Is this a property?
Regards
Shadh
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