Proving Logarithmic Equations: A Step-by-Step Guide

Hey guys! Let's dive into the exciting world of logarithms and learn how to prove some cool equations. We're going to tackle two problems that look a bit tricky at first, but don't worry, we'll break them down step by step. So grab your thinking caps, and let's get started!

Part a) Proving $a = \frac{1}{81} \sqrt[3]{9}$ if ${}^a\log 3 = -0.3$

Okay, so the first part of our challenge is to prove that if ${}^a\log 3 = -0.3$, then $a$ actually equals $\frac{1}{81} \sqrt[3]{9}$. Sounds like a mouthful, right? But trust me, it's totally doable! The key here is to really understand what a logarithm means. Remember, logarithms are just the inverse of exponents. So, ${}^a\log 3 = -0.3$ is just another way of saying $a^{-0.3} = 3$. Got it?

So, let's start with the given equation:

${}^a\log 3 = -0.3$

Now, let's convert this logarithmic equation into its exponential form. This is where the magic happens! We rewrite the equation as:

$a^{-0.3} = 3$

This equation is way easier to work with, isn't it? But we're not quite there yet. We need to isolate a. First things first, let’s get rid of that pesky decimal in the exponent. We know that -0.3 is the same as -3/10, so we can rewrite the equation as:

$a^{-\frac{3}{10}} = 3$

Now, to get a by itself, we need to raise both sides of the equation to the power of -10/3. Why -10/3? Because when you multiply the exponents (-3/10) and (-10/3), you get 1, and $a^1$ is just a! This is super important to remember when dealing with exponents and logarithms.

$(a^{-\frac{3}{10}})^{-\frac{10}{3}} = 3^{-\frac{10}{3}}$

This simplifies to:

$a = 3^{-\frac{10}{3}}$

We're getting closer! Now, let's rewrite the right side of the equation to look more like our target answer, which is $\frac{1}{81} \sqrt[3]{9}$. Remember, a negative exponent means we take the reciprocal, and a fractional exponent means we're dealing with roots. So, we can rewrite $3^{-\frac{10}{3}}$ as:

$a = \frac{1}{3^{\frac{10}{3}}}$

Now, let's break down that exponent a little further. We can rewrite $\frac{10}{3}$ as $3 + \frac{1}{3}$, so we have:

$a = \frac{1}{3^{3 + \frac{1}{3}}}$

Using the properties of exponents, we can separate this into:

a = \frac{1}{3^3 ullet 3^{\frac{1}{3}}}

We know that $3^3$ is 27, so we have:

a = \frac{1}{27 ullet 3^{\frac{1}{3}}}

Now, let's work on getting that 81 in the denominator. We know that 27 is 3 times 9, and we want something involving 81 (which is 9 times 9). Hmmm... let's rewrite $3^{\frac{1}{3}}$ as $\sqrt[3]{3}$.

$a = \frac{1}{27 \sqrt[3]{3}}$

To get closer to our target, we need to somehow introduce a factor of 3 in the cube root. Notice that $9 = 3^2$, so $\sqrt[3]{9}$ can be expressed in terms of $\sqrt[3]{3}$. Let's multiply the numerator and denominator by $\sqrt[3]{3^2}$ to rationalize the denominator.

a = \frac{1}{27 \sqrt[3]{3}} ullet \frac{\sqrt[3]{3^2}}{\sqrt[3]{3^2}}

This gives us:

a = \frac{\sqrt[3]{9}}{27 ullet 3}

Which simplifies to:

$a = \frac{\sqrt[3]{9}}{81}$

And finally, we can rewrite this as:

$a = \frac{1}{81} \sqrt[3]{9}$

Boom! We did it! We've successfully proven that if ${}^a\log 3 = -0.3$, then $a = \frac{1}{81} \sqrt[3]{9}$. Give yourself a pat on the back!

Part b) Proving $a = \frac{1}{2}\sqrt{2}$ if ${}^{\frac{1}{2}}\log (3a - \sqrt{2}) = -\frac{1}{2}$

Now, let's move on to the second part of our logarithmic adventure. This time, we need to prove that if ${}^{\frac{1}{2}}\log (3a - \sqrt{2}) = -\frac{1}{2}$, then $a = \frac{1}{2}\sqrt{2}$. The same strategy applies here: we'll start by converting the logarithmic equation into its exponential form and then solve for a. Are you ready?

Let's begin with the given equation:

${}^{\frac{1}{2}}\log (3a - \sqrt{2}) = -\frac{1}{2}$

Just like before, we convert this logarithmic equation to its exponential form:

$\left(\frac{1}{2}\right)^{-\frac{1}{2}} = 3a - \sqrt{2}$

This looks a bit more manageable already. Let's simplify the left side of the equation. Remember that a negative exponent means we take the reciprocal, and a fractional exponent of 1/2 means we're taking the square root. So, $\left(\frac{1}{2}\right)^{-\frac{1}{2}}$ is the same as $\sqrt{2}$.

$\sqrt{2} = 3a - \sqrt{2}$

Now, we need to isolate a. Let's add $\sqrt{2}$ to both sides of the equation:

$\sqrt{2} + \sqrt{2} = 3a$

This simplifies to:

$2\sqrt{2} = 3a$

Finally, to get a by itself, we divide both sides by 3:

$a = \frac{2\sqrt{2}}{3}$

Okay, we're almost there. Our target answer is $a = \frac{1}{2}\sqrt{2}$, but we have $a = \frac{2\sqrt{2}}{3}$. They look a bit different, don't they? But let's see if we can manipulate our answer to match the target.

Actually, it seems like there was a minor miscalculation in the initial steps. Let’s go back and check our work. Specifically, let's look at the step where we converted the logarithmic equation to exponential form:

${}^{\frac{1}{2}}\log (3a - \sqrt{2}) = -\frac{1}{2}$ should convert to $\left(\frac{1}{2}\right)^{-\frac{1}{2}} = 3a - \sqrt{2}$. This is correct.

$\sqrt{2} = 3a - \sqrt{2}$ is also correct.

Adding $\sqrt{2}$ to both sides gives us $2\sqrt{2} = 3a$. This is also correct.

Dividing by 3, we get $a = \frac{2\sqrt{2}}{3}$. This is still correct.

It looks like our derived answer, $a = \frac{2\sqrt{2}}{3}$, is indeed the correct simplified answer based on the given equation. The target answer of $a = \frac{1}{2}\sqrt{2}$ might be incorrect, or there might be a typo in the original problem.

However, to still try and manipulate our answer to look like the target, even though it's not mathematically equivalent, we could try a slightly unconventional approach:

We want to somehow get a 2 in the denominator. Let's multiply the numerator and denominator by something that will help us achieve this. If we multiply by $\frac{\sqrt{2}}{\sqrt{2}}$, we get:

a = \frac{2\sqrt{2}}{3} ullet \frac{\sqrt{2}}{\sqrt{2}} = \frac{2 ullet 2}{3\sqrt{2}} = \frac{4}{3\sqrt{2}}

This still doesn't look like our target. So, let's stop trying to force it. The important thing is that we followed the correct mathematical steps and arrived at a simplified answer. Sometimes, the given answer in a problem might be incorrect, and it's crucial to trust your process and calculations.

So, our final answer, based on the correct steps, is $a = \frac{2\sqrt{2}}{3}$.

Key Takeaways and Tips for Logarithmic Proofs

Okay, guys, we've tackled two pretty challenging logarithmic problems! Let's recap the key things we learned and some tips for tackling these types of problems in the future:

1. Understand the Definition of Logarithms: This is the most important thing. Logarithms are just the inverse of exponents. If you truly understand this relationship, converting between logarithmic and exponential forms will become second nature.
2. Convert to Exponential Form: When you're trying to prove logarithmic equations, the first step should almost always be to convert the logarithmic equation into its equivalent exponential form. This makes the equation much easier to manipulate and solve.
3. Master Exponent Rules: A solid understanding of exponent rules is absolutely crucial for working with logarithms. Remember things like:
   • $x^{-n} = \frac{1}{x^n}$ (negative exponents)
   • $x^{\frac{1}{n}} = \sqrt[n]{x}$ (fractional exponents)
   • x^{m+n} = x^m ullet x^n (product of powers)
   • $(x^m)^n = x^{mn}$ (power of a power)
4. Isolate the Variable: Your goal is always to get the variable you're solving for (in our case, a) by itself on one side of the equation. Use algebraic manipulations like adding, subtracting, multiplying, dividing, and raising both sides to a power to achieve this.
5. Simplify and Rationalize: Once you've isolated the variable, simplify the expression as much as possible. This might involve rationalizing denominators (getting rid of square roots in the denominator) or combining like terms.
6. Don't Be Afraid to Go Back and Check: If you're not getting the answer you expect, don't be afraid to go back and carefully check each step of your work. It's easy to make a small mistake, especially with exponents and fractions.
7. Trust Your Calculations (But Be Open to Errors): As we saw in part (b), sometimes the given answer in a problem might be incorrect. If you've carefully followed the correct steps and arrived at a different answer, trust your calculations. However, always double-check your work to be sure! It's a balancing act between confidence and diligence.

Practice Makes Perfect!

Logarithmic equations can seem intimidating at first, but with practice, you'll become a pro at solving them. The more problems you work through, the more comfortable you'll become with the different techniques and strategies. So, keep practicing, and don't give up! You've got this!

I hope this step-by-step guide has been helpful, guys! If you have any questions, feel free to ask. Happy logarithm-ing!