Solving Exponential Equations: Find The Value Of P

\cdot 10^{p+6}-6=-23$

Let's break down how to solve the equation $-9 \cdot 10^{p+6}-6=-23$ step-by-step. This involves isolating the exponential term and then using logarithms to find the value of $p$. Understanding exponential equations is crucial in various fields like finance, physics, and computer science, making this a valuable skill to acquire.

Step-by-Step Solution

1. Isolate the Exponential Term

First, we want to isolate the term that contains the variable $p$, which is $-9 \cdot 10^{p+6}$. To do this, we'll add 6 to both sides of the equation:

$-9 \cdot 10^{p+6} - 6 + 6 = -23 + 6$

This simplifies to:

$-9 \cdot 10^{p+6} = -17$

Next, we'll divide both sides by -9 to completely isolate the exponential term:

$10^{p+6} = \frac{-17}{-9}$

Which simplifies to:

$10^{p+6} = \frac{17}{9}$

2. Apply Logarithms

Now that we have the exponential term isolated, we can use logarithms to solve for $p$. Since the base of our exponential term is 10, we'll use the common logarithm (base 10), denoted as $log_{10}$ or simply $log$. Applying the logarithm to both sides of the equation gives us:

$log(10^{p+6}) = log( \frac{17}{9})$

Using the property of logarithms that $log_b(b^x) = x$, we can simplify the left side:

$p + 6 = log( \frac{17}{9})$

3. Solve for p

To solve for $p$, we simply subtract 6 from both sides of the equation:

$p = log( \frac{17}{9}) - 6$

Now, we can use a calculator to find the value of $log( \frac{17}{9})$:

$log( \frac{17}{9}) \approx 0.2745$

So,

$p \approx 0.2745 - 6$

$p \approx -5.7255$

Therefore, the value of $p$ that satisfies the equation is approximately -5.7255.

Understanding Exponential Equations

Exponential equations are equations in which the variable appears in the exponent. They are used to model various phenomena, such as population growth, radioactive decay, and compound interest. The general form of an exponential equation is $a \cdot b^{cx+d} = k$, where $a$, $b$, $c$, $d$, and $k$ are constants, and $x$ is the variable. Solving exponential equations often involves isolating the exponential term and then using logarithms.

Key Properties of Exponents and Logarithms

• Product of Powers: $a^{m} \cdot a^{n} = a^{m+n}$
• Quotient of Powers: $\frac{a^{m}}{a^{n}} = a^{m-n}$
• Power of a Power: $(a^{m})^{n} = a^{mn}$
• Logarithm of a Product: $log_b(mn) = log_b(m) + log_b(n)$
• Logarithm of a Quotient: $log_b( \frac{m}{n}) = log_b(m) - log_b(n)$
• Logarithm of a Power: $log_b(m^{n}) = n \cdot log_b(m)$
• Change of Base Formula: $log_b(a) = \frac{log_c(a)}{log_c(b)}$

Common Mistakes to Avoid

• Incorrectly Applying Logarithms: Make sure to apply logarithms correctly and use the appropriate properties. A common mistake is to try to distribute logarithms over addition or subtraction, which is not valid.
• Forgetting to Isolate the Exponential Term: Always isolate the exponential term before applying logarithms. This ensures that you are applying the logarithm to the entire exponential expression.
• Rounding Errors: Be mindful of rounding errors, especially when using a calculator. Round your answer to an appropriate number of decimal places based on the context of the problem.

Practical Applications

Exponential equations and logarithms are used in a wide range of practical applications. Here are a few examples:

Finance

In finance, exponential equations are used to calculate compound interest. The formula for compound interest is:

$A = P(1 + \frac{r}{n})^{nt}$

Where:

• $A$ is the future value of the investment/loan, including interest
• $P$ is the principal investment amount (the initial deposit or loan amount)
• $r$ is the annual interest rate (as a decimal)
• $n$ is the number of times that interest is compounded per year
• $t$ is the number of years the money is invested or borrowed for

Logarithms can be used to solve for the interest rate or the time it takes for an investment to reach a certain value.

Physics

In physics, exponential equations are used to model radioactive decay. The formula for radioactive decay is:

$N(t) = N_0 e^{-\lambda t}$

Where:

• $N(t)$ is the amount of the substance remaining after time $t$
• $N_0$ is the initial amount of the substance
• $\lambda$ is the decay constant
• $t$ is the time

Logarithms can be used to solve for the decay constant or the time it takes for a substance to decay to a certain level.

Computer Science

In computer science, logarithms are used in the analysis of algorithms. For example, the time complexity of binary search is $O(log n)$, where $n$ is the number of elements in the array. This means that the number of operations required to find an element in the array grows logarithmically with the size of the array.

Advanced Techniques

Using Natural Logarithms

While we used the common logarithm (base 10) in our example, we could also use the natural logarithm (base $e$), denoted as $ln$. The natural logarithm is often used in calculus and other advanced mathematical topics. To use the natural logarithm, we would apply it to both sides of the equation:

$ln(10^{p+6}) = ln( \frac{17}{9})$

Using the property of logarithms that $ln(e^x) = x$, we can simplify the left side:

$(p+6)ln(10) = ln( \frac{17}{9})$

Then, we can solve for $p$:

$p = \frac{ln( \frac{17}{9})}{ln(10)} - 6$

This will give us the same result as using the common logarithm.

Solving More Complex Exponential Equations

Some exponential equations may require more advanced techniques, such as substitution or factoring. For example, consider the equation:

$4^x - 2^{x+1} - 8 = 0$

We can rewrite this equation as:

$(2^x)^2 - 2 \cdot 2^x - 8 = 0$

Let $y = 2^x$. Then the equation becomes:

$y^2 - 2y - 8 = 0$

This is a quadratic equation that we can solve for $y$:

$(y - 4)(y + 2) = 0$

So, $y = 4$ or $y = -2$. Since $y = 2^x$, we have:

$2^x = 4$ or $2^x = -2$

The first equation gives us $x = 2$. The second equation has no real solutions, since $2^x$ is always positive. Therefore, the only solution to the original equation is $x = 2$.

In conclusion, solving exponential equations involves isolating the exponential term, applying logarithms, and then solving for the variable. Understanding the properties of exponents and logarithms is crucial for solving these equations. With practice, you can master these techniques and apply them to various real-world problems.

For further reading on exponential equations and logarithms, you can visit Khan Academy's page on Exponential and Logarithmic Functions.