Logaritmo #1
If lo g 2 = 0 . 3 , lo g 1 0 = 1 and lo g h = lo g ( 2 2 ) + 2 lo g 5 ,
what is the value of h ?
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5 solutions
here the question is wrong..the base could be 10 or e . so if base is 10 then its answer is 100, otherwise it is not.
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In the question you're told that log 10 = 1. If you represent the base by a variable "b", you know that b^1 = 10, which is the same as b = 10.
when no base is specified for a log ... 10 is taken as the default base
Good one..
i have a small doubt. could u plz clarify it. log 2 square=2 log2 = 2 0.3 = 0.6. log 10 = log(2 5) = log(2) + log(5) implies 0.3 +log(5) = 1. then log(5) = 0.7. so log(h) = 2 (0.3+.07) = 2. so h = 2 power 10 = 1024. cant we do the problem in this way?
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@jaswanth varma You are doing it correct. But, in the end you solved lo g 1 0 h = 2 ( 0 . 3 + 0 . 7 ) lo g 1 0 h = 2 .
But, basically, I f lo g b a = c , t h e n i t m e a n s t h a t b c = a . S o , i n t h i s c a s e 1 0 2 = h , a n d n o t 2 1 0
brother, one small correction log(25)=2*log(5) and not log(2)+log(5 )
log h = 2.log 2 + 2.log 5
log h = log 4 + log 25
log h = log( 4*25)
log h = log (100)
therefore h = 100
EASY WAY NICE
good way easy way
log h = log (4*25) log h =log (100) So h = 100
So simple, Log4+log25=log100,now,h=100
log(h)=2.log2+2log5 =2(log2+log5)=2log10=log100
l o g h = l o g 2 2 + 2 . l o g 5 l o g h = 2 . l o g 2 + 2 . l o g 5 l o g h = 2 { l o g 2 + l o g 5 } l o g h = 2 { l o g 1 0 } [ U s i n g l o g a + l o g b = l o g a b ] l o g h = 2 h = 1 0 0
Cheers!!:):)