log3 [ 9 sqrt(3) ]= log3 [ 9 ] + log3 [ sqrt(3) ]= log3 [ 32 ] + log3 [ 31/2 ]= 2 + 1/2= 2.5
log33+log3x +2=3 log33+log3x=1 log3(3x)=1 3x=3 x=1 Other interpretation: log33+log3(x+2)=3 log3(3(x+2))=3 3(x+2)=27 x+2=9 x=7
log3(x) + log9(x^3)=0 for any t >0 logt(x) = ln(x)/ln(t) so log9(x) = ln(x) / ln (9) = ln (x) / ( 2 * ln 3) = log3(x) /2 or log3(x) = 2 log9(x) log9(x^n) = n * log9(x) So log3(x) + log9(x^3) = log3(x) + 6 log3(x) = 7 log3(x) 7 log3(x) = 0 => log3(x) = 0 => x = 1
[(log1)/2](8)=(0/2)(8)=0
You need to solve the equation:log3(x-1) = 0 Taking antilogarithms (base 3) on both sides, you get: 3^log3(x-1) = 3^0 x-1 = 1 x = 2
log3 [ 9 sqrt(3) ]= log3 [ 9 ] + log3 [ sqrt(3) ]= log3 [ 32 ] + log3 [ 31/2 ]= 2 + 1/2= 2.5
use a calculator :)
you do 450 over 100 times 870 over 1 the answer you get just simplify. :)
log1 + log2 + log3 = log(1*2*3) = log6
log33+log3x +2=3 log33+log3x=1 log3(3x)=1 3x=3 x=1 Other interpretation: log33+log3(x+2)=3 log3(3(x+2))=3 3(x+2)=27 x+2=9 x=7
g/2 x 8
log3 81 × log2 8 × log4 2 = log3 (33) × log2 (23) × log4 (40.5) = 3 × (log3 3) × 3 × (log2 2) × 0.5 × (log4 4) = 3 × 1 × 3 × 1 × 0.5 × 1 = 9 × 0.5 = 4.5
It simplify to 1
Yes you can simplify it and here is the answer 5/100= 1/20 because 5 times 20 equals 5!
log3(x) + log9(x^3)=0 for any t >0 logt(x) = ln(x)/ln(t) so log9(x) = ln(x) / ln (9) = ln (x) / ( 2 * ln 3) = log3(x) /2 or log3(x) = 2 log9(x) log9(x^n) = n * log9(x) So log3(x) + log9(x^3) = log3(x) + 6 log3(x) = 7 log3(x) 7 log3(x) = 0 => log3(x) = 0 => x = 1
if you start getting a boner then its ok because like boners!
What is 7 times 1/2