Certainly! Let's work through the given equation step-by-step to find the solution:
[tex]\[ 3^x + 3^{x+1} + 3^{x+2} + 3^{x+4} = 846 \][/tex]
First, observe that [tex]\(3^{x+1}\)[/tex], [tex]\(3^{x+2}\)[/tex], and [tex]\(3^{x+4}\)[/tex] can be expressed in terms of [tex]\(3^x\)[/tex]:
[tex]\[ 3^{x+1} = 3^x \cdot 3 \][/tex]
[tex]\[ 3^{x+2} = 3^x \cdot 3^2 = 3^x \cdot 9 \][/tex]
[tex]\[ 3^{x+4} = 3^x \cdot 3^4 = 3^x \cdot 81 \][/tex]
Now, substitute these expressions back into the original equation:
[tex]\[ 3^x + 3^x \cdot 3 + 3^x \cdot 9 + 3^x \cdot 81 = 846 \][/tex]
Factor out [tex]\(3^x\)[/tex] from each term on the left side of the equation:
[tex]\[ 3^x (1 + 3 + 9 + 81) = 846 \][/tex]
Next, we need to simplify the expression inside the parentheses:
[tex]\[ 1 + 3 + 9 + 81 = 94 \][/tex]
So the equation now becomes:
[tex]\[ 3^x \cdot 94 = 846 \][/tex]
We can solve for [tex]\(3^x\)[/tex] by dividing both sides of the equation by 94:
[tex]\[ 3^x = \frac{846}{94} \][/tex]
Now, simplify the fraction:
[tex]\[ 3^x = 9 \][/tex]
Recall that 9 is a power of 3:
[tex]\[ 9 = 3^2 \][/tex]
So, we can write:
[tex]\[ 3^x = 3^2 \][/tex]
Since the bases are the same, the exponents must be equal:
[tex]\[ x = 2 \][/tex]
Thus, the solution to the equation [tex]\(3^x + 3^{x+1} + 3^{x+2} + 3^{x+4} = 846\)[/tex] is:
[tex]\[ x = 2 \][/tex]
