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Zorluk: OrtaLogarithmic and Exponential Expressions and Equations

If xx is a real number such that log⁡4(x−3)=12+log⁡4(2)\log_4(x - 3) = \frac{1}{2} + \log_4(2), what is the value of xx?

Cevap: 7

Cevap

The value of xx is 7.
By moving the logarithmic terms to the same side and applying the quotient rule, the equation simplifies to log⁡4(x−32)=12\log_4\left(\frac{x-3}{2}\right) = \frac{1}{2}. Converting this to exponential form yields x−32=41/2\frac{x-3}{2} = 4^{1/2}. Since 41/2=24^{1/2} = 2, we have x−32=2\frac{x-3}{2} = 2, which gives x−3=4x - 3 = 4, or x=7x = 7. Substituting x=7x = 7 back into the original equation shows it is a valid solution.

Adım Adım Çözüm

1
Group logarithmic terms on one side of the equation.
log⁡4(x−3)−log⁡4(2)=12\log_4(x - 3) - \log_4(2) = \frac{1}{2}
Grouping the logarithms allows them to be combined using logarithmic properties.
2
Apply the quotient property of logarithms.
log⁡4(x−32)=12\log_4\left(\frac{x - 3}{2}\right) = \frac{1}{2}
The quotient property states that log⁡b(a)−log⁡b(c)=log⁡b(ac)\log_b(a) - \log_b(c) = \log_b(\frac{a}{c}).
3
Convert the equation from logarithmic form to exponential form.
x−32=41/2\frac{x - 3}{2} = 4^{1/2}
A logarithmic equation log⁡b(y)=z\log_b(y) = z is equivalent to bz=yb^z = y.
4
Evaluate the fractional exponent and solve the linear equation for xx.
x=7x = 7
Since 41/2=24^{1/2} = 2, the equation becomes x−32=2\frac{x - 3}{2} = 2. Multiplying both sides by 2 gives x−3=4x - 3 = 4, so adding 3 to both sides yields x=7x = 7.

Anahtar Kavram

Solving logarithmic equations using properties of logarithms
Tahmini Süre:1m 30s
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