Math, asked by rajusetu, 1 year ago

llok at the attachement and solve them

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Answered by kvnmurty
1
X=12^{12}=(2^2*3)^{12}=2^{24}*3^{12}\\\\Log_{10}\ X=24\ Log_{10}\ 2+12\ Log_{10}\ 3\\.\ \ \ =24*0.3010+12*0.4771= 12.95\\\\So\ there\ are\ 13\ digits.,\ \ as\ 12 < Log_{10}\ X \le 13

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given  x³ + y³ + z³ = 3 xyz

(x+y+z)³
    = (x+y)³ + z³ + 3 (x+y)z (x+y+z)
    = (x³ + y³ + z³) + 3 x²y + 3 xy² + 3 x²z + 6 xyz + 3 xz² + 3 y²z + 3 yz²

replace the 6xyz term with 2 x³ + 2 y³ + 2 z³.  Then rearrange the terms.

    = 3 x³ + 3 y³ + 3 z³ + 3 x²y + 3 xy² + 3 x²z + 3 xz² + 3 y²z + 3 yz²
Bring terms with x² together.  those with y² together.
   = 3 x² (x+y+z) + 3 y² (y + x +z) + 3 z² (z + x +y)
   = 3 (x + y + z) (x² + y² + z²)

 x+y+z ≠ 0  as x,y,z>0 as log x, log y, log z exist.

     hence,   (x + y + z)² = 3 (x² + y² + z²)      ---- (2)

simplify the above expression by expansion to get:
               x² + y² + z² =  xy + yz + xz
                                = (x+y+z)² / 3
                


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19)  Here the logarithms are to the base 10.

Given number is expressed as    N = x * 10^y   like:   1.5015 * 10^5.

N = number.         Log               Characteristic                  Mantissa
N = x * 10^y       Log  N            Floor[ Log  N ]           Log N  -  characteristic
                       y + Log  x                 y                           Log  x

Given, N = 0.00203 = 2.03 * 10^-3,     and     Log  N  = -3 + log  2.03

So characteristic is the exponent of 10:   -3.      Mantissa = Log  2.03

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20)
      N = Log 3 = 0.4771  = 4.771 * 10^-1
      Hence :    characteristic is   -1      and the mantissa is   Log  4.771
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21)
 
Log x =  2 Log a  + 3 Log b - 4 Log c
         =  Log a² + log b³ - log c⁴
         =  Log  [ a² b³ / c⁴ ]
       so  x = a² b³ / c⁴
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22)
  Here all the logarithms are to the base 1/2.  Since all terms are for the same base, we dont need to write it.

     Log x = Log  4 + Log 3 - Log 2
               = Log  (4 * 3 / 2)
               = Log  6
         x = 6


kvnmurty: i could not do that. perhaps, log x+log y + log z = log(x^3+y^3+z^3) - log 3
kvnmurty: If this is not the answer, then check for a formula with x^3+y^3+z^3 and 3 xyz .. and then use that.
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