Solve For X: 4^x = (1/8)^(x+5)

Jan 12, 2026 by Andrew McMorgan 31 views

Hey guys! Today we're diving deep into the awesome world of exponential equations. We've got a juicy one for you: for what value of $x$ does $4^x = ( rac{1}{8})^{x+5}$ ? This might look a little intimidating at first, but trust me, by the end of this article, you'll be a pro at cracking these kinds of problems. We're going to break it down step-by-step, making sure you understand why we do each part, not just what to do. So grab your favorite beverage, get comfy, and let's get this math party started!

Understanding the Basics of Exponential Equations

Alright, let's kick things off by getting our heads around what we're dealing with here. An exponential equation is basically an equation where the variable, in this case, our $x$ , is in the exponent. Think of it like this: instead of finding a number that's multiplied by itself a certain number of times, we're trying to find the number of times something needs to be multiplied by itself to get a certain result. Pretty neat, right? The key to solving most exponential equations, especially the ones where the bases look different but are related, is to get them to a common base. This is like speaking the same language so you can have a proper conversation. In our problem, we have $4^x$ and $( rac{1}{8})^{x+5}$ . On the surface, 4 and $rac{1}{8}$ don't seem to have much in common. But if you've been around the math block a few times, you'll notice they're both related to the number 2. Yep, 2 is our secret weapon here! We can rewrite 4 as $2^2$ and $rac{1}{8}$ as $2^{-3}$ . Once we have that common base, the exponents become much easier to manage, and we can start isolating our variable $x$ . Remember, the goal is always to simplify, simplify, simplify!

Rewriting the Bases to a Common Base

So, the first major step in solving our equation $4^x = ( rac{1}{8})^{x+5}$ is to express both sides with the same base. As we hinted at before, our common base is going to be 2. Let's break down how we do that:

The Left Side: We have $4^x$ . Since $4 = 2 imes 2 = 2^2$ , we can substitute $2^2$ for 4. This gives us $(2^2)^x$ . Now, we use a cool exponent rule: when you have an exponent raised to another exponent, you multiply them. So, $(2^2)^x$ becomes $2^{2x}$ . Easy peasy!
The Right Side: Now for the trickier part, $( rac{1}{8})^{x+5}$ . First, let's deal with the $rac{1}{8}$ . Remember that a fraction like $rac{1}{a^n}$ is the same as $a^{-n}$ . So, $rac{1}{8}$ can be written as $rac{1}{2^3}$ . Applying the rule we just mentioned, $rac{1}{2^3}$ is equal to $2^{-3}$ . Awesome! Now we substitute this back into our right side: $(2^{-3})^{x+5}$ . Again, we have an exponent raised to another exponent, so we multiply them. This means we need to multiply $-3$ by the entire expression $(x+5)$ . Be careful with your signs here, guys! $-3 imes (x+5)$ becomes $-3x - 15$ . So, the right side simplifies to $2^{-3x-15}$ .

Now our original equation, $4^x = ( rac{1}{8})^{x+5}$ , looks much friendlier. After our base rewriting magic, it's transformed into $2^{2x} = 2^{-3x-15}$ . See? That common base of 2 makes all the difference!

Equating the Exponents

Okay, so we've successfully rewritten our equation with a common base: $2^{2x} = 2^{-3x-15}$ . This is where the real problem-solving fun begins! The fundamental rule here is that if you have two equal exponential expressions with the same base, then their exponents must also be equal. Think about it: if $2^a = 2^b$ , then $a$ has to equal $b$ , right? There's no other way for those two powers of 2 to be the same. This principle is our golden ticket to finding $x$ .

Since both sides of our equation are powers of 2, we can now ditch the bases and just focus on the exponents. We set the exponent from the left side equal to the exponent from the right side:

$2x = -3x - 15$

Look at that! We've gone from a complicated exponential equation to a simple linear equation. This is exactly what we wanted. Linear equations are generally much easier to solve. Our goal now is to isolate $x$ on one side of the equation. We want to get all the terms with $x$ together and all the constant terms on the other side. Let's do this!

Solving the Linear Equation for x

We're at the home stretch, guys! We have the linear equation $2x = -3x - 15$ , and our mission is to find the value of $x$ . The strategy is to gather all the $x$ terms on one side and the constant numbers on the other. This usually involves a bit of addition and subtraction.

Combine the x terms: To get all the $x$ terms on the left side, we need to get rid of the $-3x$ on the right. We can do this by adding $3x$ to both sides of the equation. Remember, whatever you do to one side, you must do to the other to keep the equation balanced.

$2x + 3x = -3x - 15 + 3x$

This simplifies to:

$5x = -15$
Isolate x: Now we have $5x = -15$ . To get $x$ by itself, we need to undo the multiplication by 5. We do this by dividing both sides of the equation by 5.

$rac{5x}{5} = rac{-15}{5}$

And voilà! This gives us:

$x = -3$

So, the value of $x$ that satisfies the original equation $4^x = ( rac{1}{8})^{x+5}$ is -3. Pretty awesome, right? We took a complex exponential problem and turned it into a straightforward linear one!

Verifying the Solution

It's always a good idea, especially in math, to verify your solution. This means plugging the value of $x$ you found back into the original equation to make sure both sides are indeed equal. It's like double-checking your work to make sure you didn't make any silly mistakes along the way. Let's plug in $x = -3$ into $4^x = ( rac{1}{8})^{x+5}$ :

Left Side: $4^x = 4^{-3}$ . To calculate this, remember that $a^{-n} = rac{1}{a^n}$ . So, $4^{-3} = rac{1}{4^3}$ . And $4^3 = 4 imes 4 imes 4 = 16 imes 4 = 64$ . So, the left side is $rac{1}{64}$ .
Right Side: $( rac{1}{8})^{x+5} = ( rac{1}{8})^{-3+5}$ . First, let's calculate the exponent: $-3 + 5 = 2$ . So, the right side becomes $( rac{1}{8})^2$ . To calculate this, we square both the numerator and the denominator: $( rac{1}{8})^2 = rac{1^2}{8^2} = rac{1}{64}$ .

Since the left side ( $rac{1}{64}$ ) is equal to the right side ( $rac{1}{64}$ ), our solution $x = -3$ is correct! This verification step gives us confidence in our answer and reinforces our understanding of how these exponential equations work.

Key Takeaways and Further Practice

So, what did we learn today, guys? We tackled the exponential equation $4^x = ( rac{1}{8})^{x+5}$ and found that the value of $x$ is -3. The main strategy for solving equations like this is to express both sides with a common base. Once you have that common base, you can equate the exponents and solve the resulting equation, which is often a linear one. Remember those exponent rules, especially when dealing with negative exponents and powers of powers!

Common Base Rule: If $a^m = a^n$ , then $m = n$ .
Negative Exponent Rule: $a^{-n} = rac{1}{a^n}$ .
Power of a Power Rule: $(a^m)^n = a^{m imes n}$ .

Practice makes perfect! Try these similar problems to really lock in the concept:

Solve for $x$ : $9^x = ( rac{1}{27})^{x-1}$
Solve for $y$ : $8^y = (2^{y+1})^2$
Solve for $z$ : $16^z = ( rac{1}{4})^{z+3}$

Keep practicing, and you'll be an exponential equation whiz in no time. If you found this breakdown helpful, give it a share! Happy calculating!