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solve using common base 3^(x + 3) - 3^x = 78
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\[3^{x+3}-3^{x}=78\]
Demmit! Hold on. I clicked prematurely.
From the first post, and using the properties of exponents, we can rewrite the first term as: \[3^{x}3^{3}-3^{x}=78\] Then, you can see that you can factor out a 3^x to get: \[3^{x}(3^{3}-1) = 78\] Simplify what's inside the parenthesis to get: \[3^{x}(27-1)=78\]\[3^{x}(26)=78\]Divide both sides by 26 to get: \[3^{x}=3\] Now, what value of x do you need to make this statement true? 1. Why because \[3^{1}=3\] Therefore, x = 1. The end.
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