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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
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∙ 11y ago
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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)
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 log1 plus log2 plus log3?
log1 + log2 + log3 = log(1*2*3) = log6
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
Log3 81 x log2 8 x log4 2 equals x?
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
What is x plus x plus 2 x equal 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
Is the following correct log3 divided by log2 equals 5 divided by 3 and if yes what are the steps?
To check the statement . . .(log3)/(log2)=5/3 ( ? )Get rid of the denominators:5log2=3log3Logarithmic property:log2^5=log3^3Eliminate the logarithms:2^5=3^3Simplify:32=27This is incorrect (that's your answer)==================================Answer #2:The answer to the question is: "No, that is incorrect"
How do you simplify log base 3 25 plus log base 3 4 and express using base 10 logarithms?
log325 + log34 = log3(25*4) = log3(100) = log10100/log103 = 2/log103
Are there integers a and b that satisfy the equation of log 3 to the base 2 equals a divided by b?
No, log3 to the base 2 is irrational.
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)
What is the Log 3 to the base 2?
x = log2(3) is the same as: 2x = 3 You can find it by: log3/log2 = .477/.30 = 1.59 (where log by itself assumes base 10, which most calculators and spreadsheets have built in functions)
