If \(2^{x} - 2^{x-1} = 32\), then the value of \(\dfrac{2x-1}{2x+3}\) is:
\(\dfrac{11}{15}\)
Factor \(2^{x-1}\) from the left side: \(2^{x}-2^{x-1}=2^{x-1}(2-1)=2^{x-1}\).
So \(2^{x-1}=32=2^{5}\), giving \(x-1 = 5\) and \(x = 6\).
Now \(2x-1 = 12-1 = 11\) and \(2x+3 = 12+3 = 15\).
Hence, \(\dfrac{2x-1}{2x+3}=\dfrac{11}{15}\).
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Important Questions from Surds and Indices
The value of (0.3) [{(200 - 146)/(3 × 3 × 3)} - 3] is:
The expression \(\frac{{15\left( {\sqrt {10} + \sqrt 5 } \right)}}{{\sqrt {10\;} + \sqrt {20} + \sqrt {40} - \sqrt 5 - \sqrt {80} }}\) is equal to:
Let \(x = \left( {\frac{{√ {1875} }}{{√ {3888} }} \div \frac{{√ {1200} }}{{\sqrt 768}}} \right) \times \frac{{√ {175} }}{{√ {1792} }}\) . Then √x is equal to:
If \(x = \sqrt {-\sqrt 3 + \sqrt {3 + 8\sqrt {7 + 4\sqrt 3}}}\) where x > 0, then the value of x is equal to:
What is the value of \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}−\sqrt{5}} \div \frac{\sqrt{14}+\sqrt{10}}{\sqrt{14}−\sqrt{10}}+\frac{\sqrt{10}}{\sqrt{5}}\) ?
