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Orion Wisoky

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4y ago

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14y ago

You basically have to take the antilogarithm - in this case, in base 3 - on both sides of the equation. That will make the "log3..." disappear on the left side. Antilog (base 3) means you raise 3 to the specified power.

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Q: What is log3 x equals -4?
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What does x equal in log 3 x equals 4?

log3(x)=4 x=3^4 x=81


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


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.


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


Log 3 plus logx equals 4?

log3 + logx=4 log(3x)=4 3x=10^4 x=10,000/3


How do you solve log base 3 of 7-log base3 of x equals 4?

log37 - log3x = 4 log3(7/x) = 4 7/x = 34 = 81 x = 7/81


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 10 log3 x -13?

10log3


What is the solution of the equation 3 to the x power equals 12?

3x = 12 log3(12) = 2.261859507 x = 2.261859507


What is 102a if log2 equals a and log3 equals b?

Find 102a if log2=a and log3=b B has no purpose in this question If a=log2, then 102a =102(log2)


3 to the power of x equals 50?

3^(x) = 3^(x) = 50 Take logs base 10 (Calculator) of both sides log3^(x) = log50 xlog3 = log50 x = log50/ log3 (NOT log (50/3)) On the calculator x = 1.6987.../0.47712... x = 3.5608.... 50


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