Math Problem Statement

solve for x in log (8x-3) base x - log 4 base x = 2

Solution

We are given the equation:

logx(8x3)logx(4)=2\log_x (8x - 3) - \log_x (4) = 2

Step 1: Use the properties of logarithms

We can combine the two logarithms using the property:

logb(A)logb(B)=logb(AB)\log_b (A) - \log_b (B) = \log_b \left( \frac{A}{B} \right)

Thus, the equation becomes:

logx(8x34)=2\log_x \left( \frac{8x - 3}{4} \right) = 2

Step 2: Eliminate the logarithm by rewriting the equation in exponential form

The logarithmic equation logx(A)=B\log_x (A) = B is equivalent to the exponential equation A=xBA = x^B. So, we rewrite the equation as:

8x34=x2\frac{8x - 3}{4} = x^2

Step 3: Solve the equation

Now, multiply both sides by 4 to eliminate the fraction:

8x3=4x28x - 3 = 4x^2

Rearrange the equation into standard quadratic form:

4x28x+3=04x^2 - 8x + 3 = 0

Step 4: Solve the quadratic equation

We can solve the quadratic equation using the quadratic formula. The general quadratic formula is:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

For the equation 4x28x+3=04x^2 - 8x + 3 = 0, the coefficients are:

  • a=4a = 4
  • b=8b = -8
  • c=3c = 3

Substitute these values into the quadratic formula:

x=(8)±(8)24(4)(3)2(4)x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(4)(3)}}{2(4)} x=8±64488x = \frac{8 \pm \sqrt{64 - 48}}{8} x=8±168x = \frac{8 \pm \sqrt{16}}{8} x=8±48x = \frac{8 \pm 4}{8}

Step 5: Find the two solutions

Now, solve for the two possible values of xx:

  1. x=8+48=128=32x = \frac{8 + 4}{8} = \frac{12}{8} = \frac{3}{2}
  2. x=848=48=12x = \frac{8 - 4}{8} = \frac{4}{8} = \frac{1}{2}

Step 6: Check the solutions

We need to check which of these solutions are valid by substituting them back into the original logarithmic equation.

  • For x=32x = \frac{3}{2}, we check the argument of the logarithm: 8x3=8(32)3=123=98x - 3 = 8\left(\frac{3}{2}\right) - 3 = 12 - 3 = 9 Since 9>09 > 0, this is valid.

  • For x=12x = \frac{1}{2}, we check the argument of the logarithm: 8x3=8(12)3=43=18x - 3 = 8\left(\frac{1}{2}\right) - 3 = 4 - 3 = 1 Since 1>01 > 0, this is also valid.

Final Answer:

Both solutions x=32x = \frac{3}{2} and x=12x = \frac{1}{2} are valid.

Do you need more details on any step or have further questions?


Here are some related questions to expand your understanding:

  1. What are the properties of logarithms and how do they help in solving equations?
  2. How can the quadratic formula be applied to different types of problems?
  3. What are the conditions for logarithms to be defined?
  4. How do you check for extraneous solutions in logarithmic equations?
  5. What is the relationship between logarithmic and exponential forms?

Tip: Always check the domain of the logarithmic function to ensure that the arguments inside the logarithms are positive.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Exponential Equations
Quadratic Equations

Formulas

log_b(A) - log_b(B) = log_b(A/B)
log_b(A) = B is equivalent to A = b^B
Quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a

Theorems

Properties of Logarithms
Quadratic Formula

Suitable Grade Level

Grades 9-12