Solve the exponential equation. 8 1 x + 1 = 2 7 x + 5 81^{x+1}=27^{x+5} 8 1 x + 1 = 2 7 x + 5

Here, we're asked to solve the exponential equation. And here, we have 81 to the power of x plus 1 is equal to 27 to the power of x plus 5. Now, this looks a little bit trickier in order to get the like bases on either side. Now 81, I know, is just 9 squared or 9 times 9, but 27 cannot be simplified into a power of 9 because it is 3 times 9. But looking at this, I know that 9 is just 3 squared, so 81 is really just 3 squared times 3 squared or 3 to the power of 4. And then 3 times 9, my 27, this just becomes 3 times 3 squared, which is really just 3 cubed. So I can actually rewrite these both as a base of 3. So this becomes 3 to the power of 4 to the power of x plus 1 is equal to 3 cubed to the power of x plus 5. Now I know that these have the same base, but let's go ahead and rewrite these simplified a little bit, getting those exponents all put together. Now this 4 will multiply the x and the 1, and this 3 will multiply the x and the 5, so we need to be careful here. So this becomes 3 to the power of 4x+4, and that's equal to 3 to the power of 3x+15 distributing those exponents. Now that I have the same bases and I have simplified my exponents, I can go ahead and just set these exponents equal to each other, 4x+3xplus15. So rewriting that down here, I get 4x+4 is equal to 3x+15. And again, I am left with a basic linear equation that I can solve. So I wanna pull all my x's to the same side, all of my constants to the other. So I'm gonna go ahead and move this 3 x over here and my 4 over here. So getting rid of that 3 x, I subtract 3 x from one side canceling it, which means I need to subtract 3 x from the other side. Now on this side, I can subtract 4 canceling it, which whatever I do to one side, I have to do to the other. So here I'm left with 4x minus 3x, which is just x, and that's equal to 15 minus 4, which is 11. And we're completely done here. We're left with our answer, x is equal to 11. Now if I were to plug 11 back into my original exponential equation, I would end up with a true statement. Thanks for watching and I'll see you in the next one.

0. Functions

Exponential & Logarithmic Equations

Calculus

0. Functions

Exponential & Logarithmic Equations

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Multiple Choice

Solve the exponential equation.
$81^{x+1}=27^{x+5}$

$negative 13$

$11$

$14$

$3$

Verified step by step guidance

First, recognize that both 81 and 27 can be expressed as powers of 3. Specifically, 81 is 3^4 and 27 is 3^3.

Rewrite the equation using these powers of 3: 81^{x+1} = 27^{x+5} becomes (3^4)^{x+1} = (3^3)^{x+5}.

Apply the power of a power property, which states (a^m)^n = a^{m*n}, to simplify both sides: 3^{4(x+1)} = 3^{3(x+5)}.

Since the bases are the same, you can set the exponents equal to each other: 4(x+1) = 3(x+5).

Solve the equation 4(x+1) = 3(x+5) for x by expanding both sides and then isolating x.

4:46m

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Multiple Choice

Solve the exponential equation.

$4^{x+7}=16$