If 4^{x} - 3^{x - frac{1}{2}} = 3^{x + frac{1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation: $4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$Rearranging terms: $4^{x} + 2^{2x - 1} = 3^{x + \frac{1}{2}} +

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) Given the equation: $4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$Rearranging terms: $4^{x} + 2^{2x - 1} = 3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}$Since $2^{2x - 1} = \frac{4^{x}}{2}$,we have: $4^{x} + \frac{4^{x}}{2} = 3^{x} \cdot \sqrt{3} + \frac{3^{x}}{\sqrt{3}}$$4^{x} \left(1 + \frac{1}{2}\right) = 3^{x} \left(\frac{3 + 1}{\sqrt{3}}\right)$$4^{x} \left(\frac{3}{2}\right) = 3^{x} \left(\frac{4}{\sqrt{3}}\right)$$\left(\frac{4}{3}\right)^{x} = \frac{4}{\sqrt{3}} \cdot \frac{2}{3} = \frac{8}{3\sqrt{3}}$$\left(\frac{4}{3}\right)^{x} = \left(\frac{2}{\sqrt{3}}\right)^{3} = \left(\left(\frac{2}{\sqrt{3}}\right)^{2}\right)^{\frac{3}{2}} = \left(\frac{4}{3}\right)^{\frac{3}{2}}$Comparing exponents,we get $x = \frac{3}{2}$.

If $4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$,then the value of $x$ is equal to

Similar Questions

The least number among $\sqrt[3]{4}, \sqrt[4]{5}, \sqrt[4]{7}$ and $\sqrt[3]{8}$ is:

If $3^x - 3^{x - 1} = 6$,then $x^x$ is equal to

$\sqrt {[10 - \sqrt {24} - \sqrt {40} + \sqrt {60}]} = $

Let $\frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4}$,then find the value of $A + B + C + D$.

If ${x^{x \cdot \sqrt[3]{x}}} = {(x \cdot \sqrt[3]{x})^x}$,then $x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module