Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\log _{10} 8 x-\log _{10}(1+\sqrt{x})=2\)

Short answer

The solution to the equation is \(x = 80\)

Step-by-step solution

Apply log rule

The first step is to apply the rule of logarithms that the difference of two logs is the log of the quotient. Using this, our equation becomes: \(\log _{10} \left(\frac{8x}{1+\sqrt{x}}\right) =2\)

Convert into exponential form

In the second step, we convert the logarithmic equation to an exponential equation. The equation then becomes \(10^2 = \frac{8x}{1+\sqrt{x}}\)

Solve for x

The next step is to solve this equation for x. Multiply both sides by \(1+\sqrt{x}\) to clear the denominator, then simplify and rearrange to isolate \(x\), the equation will be a quadratic one. This gives \(x = (1+\sqrt{x})(10^2)-(8x)\), which simplifies to \(100 + 10\sqrt{x} = 8x\). Rearranging terms and squaring both sides to get rid of the square root gives \(x^2-64x+6400=0\). Solving this equation with quadratic formula.

Use Quadratic Formula to solve x

Applying the quadratic formula, which is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=-64\), and \(c=6400\), gives us the roots for \(x\), \(x = 32 \pm 48\)

Validate solutions

We need to validate these solutions in the original equation. \(x = -16\) isn't a valid solution because we can't take a square root of a negative number or the log of a negative number. Hence, the only solution is \(x = 80\)

Key concepts

Logarithms

Logarithms are mathematical operations that help us solve equations involving exponential expressions with ease. A logarithm is the inverse operation of exponentiation. If you take \(b^y = x\), then \(\log_b(x) = y\). This is saying 'the power to which the base \(b\) must be raised to yield \(x\)'.
Understanding these basics is crucial when dealing with logarithmic equations. Logarithms can simplify many computations and solve equations that appear complex at first glance. For instance, the step-by-step solution of the given equation involved simplifying a quotient inside a \(\log\) by applying the rule of logarithms known as the quotient rule:

• \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)

By using this rule, we switch from an equation with two logs into one, making our calculations more manageable. Understanding these fundamental log rules is essential for solving logarithmic equations.

Exponential Form

Transforming a logarithmic equation into an exponential form can simplify its solution. This method is an integral step in solving many logarithmic problems. For an equation \(\log_b(N) = A\), transforming into exponential form gives \(b^A = N\).
In our problem, converting the logarithmic equation \(\log_{10}\left(\frac{8x}{1+\sqrt{x}}\right) = 2\) to its exponential form involves raising the base 10 to the power 2. This step transforms the original equation into a more tractable form:

• \(10^2 = \frac{8x}{1+\sqrt{x}}\) gives a clearer path to determining the variable inside the logarithm.

Shifting from logarithmic to exponential form is often a strategic move designed to simplify intricate expressions for easier manipulation, allowing us to solve for unknown variables more effectively.

Quadratic Formula

The quadratic formula is indispensable for solving quadratic equations of the form \(ax^2 + bx + c = 0\). This formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), provides solutions by using the coefficients \(a\), \(b\), and \(c\) directly.
In our exercise, after converting the logarithmic form to a quadratic equation \(x^2 - 64x + 6400 = 0\), the quadratic formula steps in to find the values of \(x\).
The coefficients are identified as:

• \(a = 1\)
• \(b = -64\)
• \(c = 6400\)

Applying the quadratic formula to these coefficients, the root calculations led to the solutions \(x = 32 \pm 48\). This versatility makes the quadratic formula a powerful tool for algebraic problem-solving.

Algebraic Solutions

Algebraic solutions refer to methods used to find the exact numbers or expressions that satisfy an equation. They often involve simplifying expressions, factoring, applying formulas, and verifying solutions.
For our loaded equation, the task was to rearrange and transform terms to solve for \(x\). After transforming into exponential and quadratic forms, as illustrated previously, a crucial algebraic operation was to verify solutions.
Upon reaching potential solutions, \(x = 80\) and \(x = -16\), verification against the original equation showed that \(x = 80\) was valid, while \(x = -16\) was discarded as it did not fit the context of the problem (negative square roots or logarithms are undefined). The algebraic approach ensures not only that solutions make mathematical sense, but also that they are practical and applicable.