What is the result of log 10(7)*log 10(343)*log 7(100)* log 7( 1000)?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to find the result of log 10(7)*log 10(343)*log 7(100)* log 7( 1000).

First we use the relation that log x^n = n log x.

log 10(7)*log 10(343)*log 7(100)* log 7( 1000)

=> log 10(7)*log 10(7^3)*log 7(10^2)* log 7( 10^3)

=> 3* log 10(7)*log 10(7)*2*3*log 7(10)* log 7( 10)

=> 18*log 10(7)*log 10(7)*log 7(10)* log 7( 10)

Now we use the relation that log a (b) = 1/ log b(a)

This gives log 7 (10) = 1/ log 10 (7)

18*log 10(7)*log 10(7)*log 7(10)* log 7( 10)

= 18 log 10(7)*log 10(7)/ log 10(7)*log 10(7)

= 18

Therefore the required result is 18.

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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 Let E = log 10(7)*log 10(343)*log 7(100)* log 7( 1000)

Let us rewrtie the numbers:

We know that :

343 = 7^3

100 = 10^2

1000 = 10^3

==> E = log 10 (7) * log 10 (7^3) *log 7 (10^2) * log 7 (10)^3

From algorethim properties we know that:

log a^b = b*log a:

==> E = log10 (7) *3log10(7) * 2log7(10) * 3log7 (10)

          = 18*log10(7) *log7(10)

We will rewrtie log7(10)

We know that:

log a b = log c b/ log c a

==> log7 (10) = log 10 10/ log 10 (7) = 1/log 10 (7)

==> E = 18*log 10 (7) *[ 1/log 10 (7)

            = 18* 1

   ==> E = 18

 

 

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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To find log10 (7)*log10(343)*log7(100)*log7(1000).

WE know that log a (b) = log c (b) }/logc (a)

Therefore log 7 (100) = {log10 (100)}/log 10 (7).

log7 (100) =   2/log10(7)....(1)

Similarly, log 7(1000) = (log 10 (1000)}/log10 (7)

log 7 (1000) = 3/log10 (7)....(2).

Using (1) and (2), we rewrite the given expression:

log 10 (7)* log 10 (343) *log 7 (100)*log7 (1000) = log 10(7)* log 10 (7^3) * {2/log10 (7)}{3/log10 (7) = log10(7)* 3log10(7)*{2/log10(7)}*{3/log10(7)} = 3*2*3 = 18.

Therefore log10(7)*log10(343)*log7(100)*log7(1000) = 18.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll write conventionally the decimal logarithms as log 10 x = lg x

We'll try to write the arguments of logarithms, except the prime numbers, as powers.

For instance:

343 = 7*7*7 = 7^3

We'll take logarithms both sides and we'll apply product rule:

lg 343 = lg(7*7*7) = lg7 + lg7 + lg7 = 3lg7

100 = 10*10 = 10^2

We'll take logarithms both sides and we'll apply product rule:

log 7 100 = log 7 (10*10) = log 7 10 + log 7 10 = 2 log 7 10

1000 = 10*10*10 = 10^3

We'll take logarithms both sides and we'll apply product rule:

log 7 1000 = log 7 (10*10*10) = log 7 10+ log 7 10 + log 7 10 = 3 log 7 10

The expression will become:

lg(7)*(3lg7)*(2 log 7 10)* (3 log 7 10) = 18[(lg 7)^2]*(log 7 10)^2

But log 7 10 = 1/lg 7

18[(lg 7)^2]*(log 7 10)^2 = 18[(lg 7)^2]/(lg 7)^2

We'll simplify and we'll get:

18[(lg 7)^2]*(log 7 10)^2 = 18

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