Solve: -1 + Log(5/3)((2x+1)/3) = -3 | Step-by-Step Solution

Hey guys! Today, we're diving into the world of logarithmic equations and tackling a specific problem that might seem a bit tricky at first. But don't worry, we'll break it down step-by-step so you can conquer these types of problems with confidence. Our mission? To solve the equation: $-1 + \log_{\frac{5}{3}}(\frac{2x+1}{3}) = -3$ for the elusive variable x. We'll not only find the solution but also understand the why behind each step. So, let's get started and unlock the secrets of logarithmic equations!

Understanding the Problem

Before we jump into solving, let's make sure we understand what we're dealing with. This equation involves a logarithm, which is essentially the inverse operation of exponentiation. The expression $\log_{\frac{5}{3}}(\frac{2x+1}{3})$ asks the question: "To what power must we raise $\frac{5}{3}$ to get $\frac{2x+1}{3}$?" The base of our logarithm is $\frac{5}{3}$, and the argument (the thing inside the logarithm) is $\frac{2x+1}{3}$. Our goal is to isolate x, but we need to first get rid of the logarithm. To do this, we'll use the properties of logarithms and exponents. We'll carefully manipulate the equation, ensuring we maintain equality at each step. Think of it like a delicate balancing act, where we need to perform the same operations on both sides to keep everything in equilibrium. So, let's put on our mathematical hats and get ready to solve this equation!

Isolating the Logarithm

The first step in solving this logarithmic equation is to isolate the logarithmic term. This means getting the $\log_{\frac{5}{3}}(\frac{2x+1}{3})$ part all by itself on one side of the equation. Currently, we have a "-1" hanging out on the left side, which we need to get rid of. The easiest way to do this is to add 1 to both sides of the equation. This is a fundamental algebraic principle: whatever you do to one side of the equation, you must do to the other to maintain balance. So, adding 1 to both sides, we get:

$-1 + \log_{\frac{5}{3}}(\frac{2x+1}{3}) + 1 = -3 + 1$

This simplifies to:

$\log_{\frac{5}{3}}(\frac{2x+1}{3}) = -2$

Now, the logarithmic term is nicely isolated on the left side. We're one step closer to unraveling the value of x. By isolating the logarithm, we've set the stage for the next crucial step: converting the logarithmic equation into its equivalent exponential form. This transformation will allow us to eliminate the logarithm altogether and work with a more familiar algebraic expression. So, with the logarithm isolated, we're ready to move on and unleash the power of exponential conversion!

Converting to Exponential Form

Now that we've isolated the logarithm, it's time to transform the equation into its exponential form. This is the key to unlocking the x inside the logarithm. Remember, logarithms and exponents are two sides of the same coin. The logarithmic equation $\log_{b}(a) = c$ is equivalent to the exponential equation $b^c = a$. In our case, we have $\log_{\frac{5}{3}}(\frac{2x+1}{3}) = -2$. So, applying the conversion rule, we can rewrite this as:

$(\frac{5}{3})^{-2} = \frac{2x+1}{3}$

This transformation is a game-changer! We've successfully eliminated the logarithm and now have a simple algebraic equation to solve. But before we dive into isolating x, let's simplify the left side of the equation. We have a fraction raised to a negative power, which might look intimidating, but it's actually quite straightforward to handle. Remember that a negative exponent means we take the reciprocal of the base and raise it to the positive exponent. So, let's conquer this negative exponent and move closer to finding our solution!

Simplifying the Exponential Term

Let's tackle that exponential term: $(\frac{5}{3})^{-2}$. As we discussed, a negative exponent tells us to take the reciprocal of the base and change the exponent to positive. The reciprocal of $\frac{5}{3}$ is $\frac{3}{5}$. So, we can rewrite the expression as:

$(\frac{5}{3})^{-2} = (\frac{3}{5})^{2}$

Now we have a positive exponent, which is much easier to deal with. Squaring a fraction means squaring both the numerator and the denominator:

$(\frac{3}{5})^{2} = \frac{3^2}{5^2} = \frac{9}{25}$

So, our equation now looks like this:

$\frac{9}{25} = \frac{2x+1}{3}$

We've successfully simplified the exponential term and transformed our equation into a more manageable form. Now we have a proportion, which is an equation stating that two ratios are equal. To solve for x, we can use the technique of cross-multiplication, which will help us eliminate the fractions and get closer to isolating x. So, let's get ready to cross-multiply and continue our journey towards the solution!

Solving for x

Now that we have the equation $\frac{9}{25} = \frac{2x+1}{3}$, we're ready to solve for x. As mentioned earlier, we can use cross-multiplication to eliminate the fractions. Cross-multiplication involves multiplying the numerator of one fraction by the denominator of the other fraction and setting the results equal. So, we multiply 9 by 3 and 25 by (2x+1):

$9 * 3 = 25 * (2x + 1)$

This gives us:

$27 = 50x + 25$

Now we have a simple linear equation. To isolate x, we first subtract 25 from both sides:

$27 - 25 = 50x + 25 - 25$

This simplifies to:

$2 = 50x$

Finally, we divide both sides by 50 to solve for x:

$\frac{2}{50} = \frac{50x}{50}$

This gives us:

$x = \frac{2}{50}$

We can simplify this fraction by dividing both the numerator and denominator by their greatest common divisor, which is 2:

$x = \frac{2 \div 2}{50 \div 2} = \frac{1}{25}$

So, the solution to our equation is $x = \frac{1}{25}$. We've successfully navigated the logarithmic equation, converted it to exponential form, simplified, and isolated x! But before we celebrate, it's always a good idea to check our answer to make sure it's valid.

Checking the Solution

It's crucial to check our solution, $x = \frac{1}{25}$, in the original equation to ensure it's valid. This is especially important with logarithmic equations because logarithms are only defined for positive arguments. If plugging in our solution results in taking the logarithm of a non-positive number, then it's an extraneous solution and we need to discard it. Let's substitute $x = \frac{1}{25}$ back into the original equation:

$-1 + \log_{\frac{5}{3}}(\frac{2(\frac{1}{25})+1}{3}) = -3$

First, let's simplify the argument of the logarithm:

$\frac{2(\frac{1}{25})+1}{3} = \frac{\frac{2}{25}+1}{3} = \frac{\frac{2}{25} + \frac{25}{25}}{3} = \frac{\frac{27}{25}}{3} = \frac{27}{25} * \frac{1}{3} = \frac{9}{25}$

Now, our equation looks like this:

$-1 + \log_{\frac{5}{3}}(\frac{9}{25}) = -3$

Let's evaluate the logarithm. We need to find the power to which we must raise $\frac{5}{3}$ to get $\frac{9}{25}$. Notice that $\frac{9}{25}$ can be written as $(\frac{3}{5})^2$, which is also $(\frac{5}{3})^{-2}$. So:

$\log_{\frac{5}{3}}(\frac{9}{25}) = -2$

Substituting this back into our equation:

$-1 + (-2) = -3$

$-3 = -3$

Our solution checks out! $x = \frac{1}{25}$ is indeed the solution to the equation. We've successfully solved the logarithmic equation and verified our answer. Give yourselves a pat on the back!

Final Answer

Therefore, the solution to the equation $-1 + \log_{\frac{5}{3}}(\frac{2x+1}{3}) = -3$ is:

D. $x = \frac{1}{25}$

Great job, guys! You've tackled a logarithmic equation and emerged victorious. Remember the steps we followed: isolate the logarithm, convert to exponential form, simplify, solve for x, and check your solution. With practice, you'll become logarithmic equation-solving pros!