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Zorluk: OrtaLogarithms and Change of Base

If log9x+log3x=3\log_9 x + \log_3 x = 3, what is the value of xx?

  1. 99Cevap
  2. B
    33
  3. C
    2727
  4. D
    8181

Cevap

The value of xx is 99.
Using the change of base formula, log9x=log3xlog39=12log3x\log_9 x = \frac{\log_3 x}{\log_3 9} = \frac{1}{2}\log_3 x. Adding log3x\log_3 x yields 32log3x=3\frac{3}{2}\log_3 x = 3, which simplifies to log3x=2\log_3 x = 2. Converting to exponential form gives x=32=9x = 3^2 = 9.

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1
Express log9x\log_9 x in base 33 using the change of base formula.
log9x=log3xlog39=log3x2=12log3x\log_9 x = \frac{\log_3 x}{\log_3 9} = \frac{\log_3 x}{2} = \frac{1}{2}\log_3 x
Converting all logarithmic terms to a common base (base 3) allows combining terms.
2
Substitute 12log3x\frac{1}{2}\log_3 x into the given equation.
12log3x+log3x=3    32log3x=3\frac{1}{2}\log_3 x + \log_3 x = 3 \implies \frac{3}{2}\log_3 x = 3
Combine like logarithmic terms.
3
Solve for log3x\log_3 x.
log3x=3×23=2\log_3 x = 3 \times \frac{2}{3} = 2
Multiply both sides by 23\frac{2}{3} to isolate log3x\log_3 x.
4
Convert from logarithmic form to exponential form.
x=32=9x = 3^2 = 9
By definition, logba=c\log_b a = c is equivalent to bc=ab^c = a.

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Logarithms and Change of Base
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