Demonstrate Logarithmic Equalities: Step-by-Step Solutions

Hey guys! Today, we're diving deep into the fascinating world of logarithms and tackling some challenging equality demonstrations. Logarithms might seem intimidating at first, but don't worry, we'll break down each problem step-by-step, making sure you understand the logic and techniques involved. Get ready to sharpen your math skills and impress your friends with your newfound logarithmic prowess!

1) Demonstrating $9^{2\log_3 2+4\log_{81} 2} \cdot \sqrt{3^2+0.5\log_3 16} = 384$

This problem looks complex, right? But let's break it down. Our main goal here is to simplify each part of the equation using logarithmic properties and then combine them to show that the left side indeed equals 384. We'll need to utilize key properties like the power rule, change of base formula, and the definition of logarithms themselves. So, buckle up, and let's get started!

Step 1: Simplify the Exponent of 9

First, let's focus on simplifying the exponent of 9: $2\log_3 2 + 4\log_{81} 2$. This is where the power rule and change of base formula come into play. The power rule of logarithms states that $a\log_b c = \log_b c^a$. The change of base formula is $\log_a b = \frac{\log_c b}{\log_c a}$, which allows us to convert logarithms from one base to another, often simplifying calculations. Our aim is to express both logarithmic terms with the same base, which in this case, appears to be 3.

Let's apply the power rule:

$2\log_3 2 = \log_3 2^2 = \log_3 4$

Now, let's tackle the second term, $4\log_{81} 2$. We'll use the change of base formula to convert the base from 81 to 3. Remember that $81 = 3^4$, which will be helpful in our conversion:

$4\log_{81} 2 = 4 \cdot \frac{\log_3 2}{\log_3 81} = 4 \cdot \frac{\log_3 2}{\log_3 3^4}$

Using the property $\log_a a^x = x$, we get:

$4 \cdot \frac{\log_3 2}{4} = \log_3 2$

Now, we can add the simplified logarithmic terms:

$\log_3 4 + \log_3 2$

Using the product rule of logarithms, which states that $\log_b m + \log_b n = \log_b (m \cdot n)$, we have:

$\log_3 4 + \log_3 2 = \log_3 (4 \cdot 2) = \log_3 8$

So, the exponent of 9 simplifies to $\log_3 8$. This is a significant step forward in making the equation easier to handle.

Step 2: Simplify the Square Root Term

Next, let's simplify the square root term: $\sqrt{3^2 + 0.5\log_3 16}$. We need to simplify the expression inside the square root first. Let's begin by simplifying $0.5\log_3 16$.

Remember that $0.5$ is the same as $\frac{1}{2}$, and $16 = 2^4$. So, we can rewrite the term as:

$\frac{1}{2} \log_3 2^4$

Using the power rule of logarithms, we get:

$\frac{1}{2} \cdot 4 \log_3 2 = 2\log_3 2$

Now, apply the power rule again:

$2\log_3 2 = \log_3 2^2 = \log_3 4$

Now, let's substitute this back into the square root expression:

$\sqrt{3^2 + \log_3 4} = \sqrt{9 + \log_3 4}$

We can express 9 as a logarithm with base 3. Since $3^2 = 9$, we can write 9 as $\log_3 3^9$. This seems counterintuitive, but it will help us combine terms:

So actually $9 = \log_3 3^9 $ is incorrect, the correct way to write it would be $9 = \log_3 3^9 = \log_3 19683$ . Let's correct the step by step calculation.

$\sqrt{9 + \log_3 4} = \sqrt{\log_3 3^9 + \log_3 4} $

$\sqrt{9 + \log_3 4} = \sqrt{\log_3 19683 + \log_3 4} $

Using the product rule of logarithms, we get:

$\sqrt{\log_3 (19683 \cdot 4)} = \sqrt{\log_3 78732}$

This form is hard to simplify further directly. Let's go back and see if we can simplify $\sqrt{9 + \log_3 4}$ in another way. We recognize that $\log_3 4$ can be approximated but it doesn't lead to a clean simplification. Let's keep this term as is for now and see how it plays out in the next steps.

Step 3: Substitute Simplified Terms Back into the Original Equation

Now that we've simplified the exponent of 9 and the square root term, let's substitute them back into the original equation:

$9^{\log_3 8} \cdot \sqrt{9 + \log_3 4} = 384$

We can rewrite $9$ as $3^2$, so the equation becomes:

$(3^2)^{\log_3 8} \cdot \sqrt{9 + \log_3 4} = 384$

Using the power of a power rule, $(a^m)^n = a^{m \cdot n}$, we have:

$3^{2\log_3 8} \cdot \sqrt{9 + \log_3 4} = 384$

Apply the power rule of logarithms to the exponent:

$3^{\log_3 8^2} \cdot \sqrt{9 + \log_3 4} = 384$

$3^{\log_3 64} \cdot \sqrt{9 + \log_3 4} = 384$

Since $a^{\log_a x} = x$, we get:

$64 \cdot \sqrt{9 + \log_3 4} = 384$

Step 4: Isolate the Square Root and Solve

Now, let's isolate the square root by dividing both sides by 64:

$\sqrt{9 + \log_3 4} = \frac{384}{64} = 6$

Square both sides of the equation:

$9 + \log_3 4 = 36$

Subtract 9 from both sides:

$\log_3 4 = 27$

This is where we seem to hit a snag. The equation $\log_3 4 = 27$ is incorrect because $3^{27}$ is an extremely large number, not 4. It indicates there might have been an error in our simplification or transcription of the original problem. Let's double-check the original equation and our steps to identify any mistakes.

Upon reviewing the steps, a mistake was found in Step 2. The intention was to write 9 as a logarithm with base 3 to combine it with the other logarithm, but the conversion was incorrect. The correct way to think about it is recognizing that we should aim for an exact value within the square root, if possible. Let’s rewind and correct that step.

Corrected Step 2: Simplify the Square Root Term (Revised)

We have $\sqrt{3^2 + 0.5\log_3 16} = \sqrt{9 + 0.5\log_3 16}$. We already simplified $0.5\log_3 16$ to $\log_3 4$. So the square root term becomes:

$\sqrt{9 + \log_3 4}$

Instead of trying to convert 9 into a logarithm with base 3 directly (which led to the error), let’s see how this term plays out in the larger equation.

Revised Step 3: Substitute Simplified Terms Back into the Original Equation

Substituting the corrected simplification back into the original equation, we have:

$9^{\log_3 8} \cdot \sqrt{9 + \log_3 4} = 384$

We already simplified $9^{\log_3 8}$ to 64. So the equation is:

$64 \cdot \sqrt{9 + \log_3 4} = 384$

Revised Step 4: Isolate the Square Root and Solve

Divide both sides by 64:

$\sqrt{9 + \log_3 4} = \frac{384}{64} = 6$

Square both sides:

$9 + \log_3 4 = 36$

Subtract 9 from both sides:

$\log_3 4 = 27$

Still, we arrive at the same incorrect equation: $\log_3 4 = 27$. This indicates a potential transcription error in the original problem statement or a more subtle error in the simplification process that we haven't yet identified. It’s crucial to go back and meticulously check each step, paying close attention to logarithmic properties and arithmetic.

Further Review and Correction

After careful review, the error lies in how we handled the initial exponent simplification. Let's revisit Step 1 with an even finer comb.

Step 1 (Revisited and Corrected): Simplify the Exponent of 9

We have $2\log_3 2 + 4\log_{81} 2$. We correctly applied the power rule to get $\log_3 4$ for the first term. For the second term, $4\log_{81} 2$, we used the change of base formula. Let's write it out meticulously again:

$4\log_{81} 2 = 4 \cdot \frac{\log_3 2}{\log_3 81}$

Since $81 = 3^4$, $\log_3 81 = 4$, so:

$4 \cdot \frac{\log_3 2}{4} = \log_3 2$

So far, so good. Now, the sum:

$\log_3 4 + \log_3 2 = \log_3 (4 \cdot 2) = \log_3 8$

This is correct. The exponent of 9 is indeed $\log_3 8$.

The error likely stems from a misinterpretation of the original equation or a small arithmetic error in the later steps. We need to meticulously re-examine every step to pinpoint the exact moment the calculation went astray.

Going back to Step 3 and Step 4 with an extremely critical eye:

We had:

$64 \cdot \sqrt{9 + \log_3 4} = 384$

Dividing both sides by 64:

$\sqrt{9 + \log_3 4} = 6$

Squaring both sides:

$9 + \log_3 4 = 36$

Subtracting 9:

$\log_3 4 = 27$

The error is not in the algebraic manipulation but in the implication. The equation $\log_3 4 = 27$ is mathematically incorrect and signals a previous error in simplifying the square root term itself. We need to dissect $\sqrt{9 + 0.5\log_3 16}$ with extra care.

Revisiting the Square Root Simplification (Again!)

We have $\sqrt{9 + 0.5\log_3 16}$. We correctly simplified $0.5\log_3 16$ to $\log_3 4$. So, we have:

$\sqrt{9 + \log_3 4}$

This is where the critical insight lies: We cannot easily combine 9 and $\log_3 4$ into a single logarithm that simplifies neatly. We made an implicit assumption that this term should simplify to a nice whole number, which is not necessarily true.

Let's pause and reconsider our approach. We've simplified as much as we can using logarithmic properties. The key is to recognize that the problem might be designed to test our understanding of how these simplifications interact within the larger equation, not necessarily to yield a perfect, clean integer at every step.

The Correct Approach: Retracing and Re-evaluating

Let's go all the way back to the original equation and work through it meticulously, step-by-step, double-checking each application of logarithmic properties and arithmetic operations.

Original Equation:

$9^{2\log_3 2+4\log_{81} 2} \cdot \sqrt{3^2+0.5\log_3 16} = 384$

We've already established that the exponent of 9 simplifies to $\log_3 8$. So, the equation becomes:

$9^{\log_3 8} \cdot \sqrt{3^2+0.5\log_3 16} = 384$

Which we rewrote as:

$64 \cdot \sqrt{9 + \log_3 4} = 384$

Here’s the critical observation: Instead of forcing the square root term to simplify further, let’s isolate it and see what we get.

Divide both sides by 64:

$\sqrt{9 + \log_3 4} = 6$

Square both sides:

$9 + \log_3 4 = 36$

Subtract 9:

$\log_3 4 = 27$

This is where we realized the inconsistency before. The problem isn't in the algebra, it's in the inherent nature of the equation itself. There's no way for $\log_3 4$ to equal 27 because $3^{27}$ is astronomically larger than 4. This strongly suggests an error in the original problem statement itself.

Conclusion for Part 1:

After thorough and repeated review, the most likely explanation is that there is an error in the original equation provided. The steps taken to simplify the equation using logarithmic properties are correct, but the resulting equation leads to an impossible condition. Therefore, we cannot demonstrate the equality as it is written.

2) Demonstrating $\log_3 2 \cdot \log_4 3 \cdot \log_5 4 \cdot \log_6 5 \cdot \log_7 6 \cdot \log_8 7$

This problem, on the other hand, looks like a classic application of the change of base formula. The key here is to strategically change the base of each logarithm so that terms cancel out, leaving us with a much simpler expression. Think of it like a chain reaction where each conversion simplifies the next one. Let's dive in!

Step 1: Apply the Change of Base Formula

The change of base formula, as we mentioned earlier, is $\log_a b = \frac{\log_c b}{\log_c a}$. We'll apply this formula to each term in the product. A smart choice for the new base here is a common one, like base 10, or even better, a base that allows for cascading cancellations. We will use the common base 10 logarithm, denoted as $\log$.

Let's apply the change of base formula to each term:

• $\log_3 2 = \frac{\log 2}{\log 3}$
• $\log_4 3 = \frac{\log 3}{\log 4}$
• $\log_5 4 = \frac{\log 4}{\log 5}$
• $\log_6 5 = \frac{\log 5}{\log 6}$
• $\log_7 6 = \frac{\log 6}{\log 7}$
• $\log_8 7 = \frac{\log 7}{\log 8}$

Step 2: Substitute and Simplify

Now, let's substitute these expressions back into the original product:

$\frac{\log 2}{\log 3} \cdot \frac{\log 3}{\log 4} \cdot \frac{\log 4}{\log 5} \cdot \frac{\log 5}{\log 6} \cdot \frac{\log 6}{\log 7} \cdot \frac{\log 7}{\log 8}$

Notice the beautiful cancellation pattern? The $\log 3$ in the first denominator cancels with the $\log 3$ in the second numerator, and so on. This creates a cascading cancellation effect, which is precisely what we were aiming for!

After all the cancellations, we are left with:

$\frac{\log 2}{\log 8}$

Step 3: Further Simplification

We can simplify this further. Recognize that $8 = 2^3$. So, we can rewrite $\log 8$ as $\log 2^3$.

Using the power rule of logarithms, $\log a^b = b \log a$, we have:

$\log 2^3 = 3 \log 2$

Substitute this back into our expression:

$\frac{\log 2}{3 \log 2}$

Now, the $\log 2$ terms cancel out, leaving us with:

$\frac{1}{3}$

Conclusion for Part 2:

Therefore, $\log_3 2 \cdot \log_4 3 \cdot \log_5 4 \cdot \log_6 5 \cdot \log_7 6 \cdot \log_8 7 = \frac{1}{3}$. We have successfully demonstrated this equality using the change of base formula and strategic cancellation.

3) Demonstrating \frac{\log_a ... }

The provided input is incomplete for the third problem. It starts with \frac{\log_a but doesn't provide the rest of the equation or the expression that needs to be demonstrated. To properly address this, we would need the full expression or equality that needs to be proven.

Final Thoughts

Logarithmic problems can be tricky, but with a solid understanding of the fundamental properties and a methodical approach, they become much more manageable. Remember to double-check your work, be on the lookout for potential simplifications, and don't be afraid to revisit your steps if you encounter an inconsistency. And most importantly, guys, keep practicing! The more you work with logarithms, the more comfortable and confident you'll become. Happy calculating!