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How do you solve Log base 3 of 7x equals Log base 3 of 2x plus Log base 3 of 0.5?
There is not a solution. Knowing how logarithms work helps. On the right hand side you have: log3(2*x) + log3(0.5).Adding logs is equivalent to multiplying (inside the log). This becomes: log3(0.5 * 2*x) = log3(x).Subtract log3(x) from both sides: log3(7x) - log3(x) = 0.Subtracting logs is equivalent to division (inside the log): log3(7x/x) = 0.So log3(7) = 0, which is Not true. No Solution.
How do you solve this log x base 3 plus log x3 base 9 equal 0?
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
What is Log3 3 plus log3 x plus 2 equals 3?
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
Simplify log3 times 1 over 3?
Log(3 * 1/3) = log(1) = 0
What is the zero of the logarithmic function f(x) log3 (x-1)?
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
