Math Problem Statement

log9y= 1 2

Solution

It seems like you're trying to solve or work with a logarithmic equation, but the expression you provided is a bit unclear. I'll assume you're referring to an equation in the form of:

log9(y)=12\log_9(y) = \frac{1}{2}

If this is correct, here's how you can solve it:

Step-by-Step Solution

  1. Rewrite the logarithmic equation in exponential form: By the definition of logarithms, logb(a)=c\log_b(a) = c is equivalent to bc=ab^c = a. So we can rewrite log9(y)=12\log_9(y) = \frac{1}{2} as:

    912=y9^{\frac{1}{2}} = y

  2. Simplify the exponential expression: 9129^{\frac{1}{2}} is the square root of 9, so:

    9=3\sqrt{9} = 3

  3. Conclusion: Therefore, y=3y = 3.


Would you like more details on this solution or have any specific questions? Here are five related questions you can explore further:

  1. What is the general form of logarithmic and exponential equations?
  2. How can logarithms with different bases be converted?
  3. What happens when you have logb(y)=c\log_b(y) = c with more complex numbers?
  4. How does solving logarithmic equations differ when there is a coefficient in front of logb(y)\log_b(y)?
  5. Can logarithms be solved graphically?

Tip: When solving logarithmic equations, remember that logb(a)=c\log_b(a) = c translates to bc=ab^c = a, which helps in converting between logarithmic and exponential forms.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Exponential Equations

Formulas

log_b(a) = c ⇔ b^c = a
√9 = 3

Theorems

Logarithmic-Exponential Relationship

Suitable Grade Level

Grades 9-12