Question - Solving Exponential Equations with Same Base
Solution:
To solve the equation $$ 8^{x+1} = 2^{5-x} $$ using a common base, we first need to express both sides of the equation with the same base. We know that 8 is equivalent to $$ 2^3 $$. Thus, we can rewrite $$ 8^{x+1} $$ as $$ (2^3)^{x+1} $$, which simplifies to $$ 2^{3(x+1)} $$.Now the equation becomes:$$ 2^{3(x+1)} = 2^{5-x} $$Since the bases are now the same, we can set the exponents equal to each other:$$ 3(x+1) = 5 - x $$Next, we distribute the 3 on the left side:$$ 3x + 3 = 5 - x $$Now bring all x terms to one side and constants to the other side:$$ 3x + x = 5 - 3 $$Combine like terms:$$ 4x = 2 $$Finally, divide by 4 to get x alone:$$ x = \frac{2}{4} $$Simplify the fraction:$$ x = \frac{1}{2} $$So the solution is x = 1/2.
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