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What is log3 9root3?
log3 [ 9 sqrt(3) ]= log3 [ 9 ] + log3 [ sqrt(3) ]= log3 [ 32 ] + log3 [ 31/2 ]= 2 + 1/2= 2.5
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 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
What is log base 3 of (x plus 1) log base 2 of (x-1)?
The browser which is used for posting questions is almost totally useless for mathematical questions since it blocks most symbols.I am assuming that your question is about log base 3 of (x plus 1) plus log base 2 of (x-1).{log[(x + 1)^log2} + {log[(x - 1)^log3}/log(3^log2) where all the logs are to the same base - whichever you want. The denominator can also be written as log(3^log2)This can be simplified (?) to log{[(x + 1)^log2*(x - 1)^log3}/log(3^log2).As mentioned above, the expression can be to any base and so the expression becomesin base 2: log{[(x + 1)*(x - 1)^log3}/log(3) andin base 3: log{[(x + 1)^log2*(x - 1)}/log(2)
