Solve for $x$ $$4^x-\dfrac{3^x}{\sqrt3}=3^x\cdot\sqrt3-\dfrac{2^{2x}}2$$ I don't understand how to convert it into quadratic equation how should I equate all bases
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can you use $\LaTeX$ please? – Dr. Sonnhard Graubner Jun 26 '15 at 16:06
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Nothing quadratic here. – Did Jun 26 '15 at 17:51
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As $2^{2x}=(2^2)^x=4^x,$
$$4^x-\dfrac{3^x}{\sqrt3}=3^x\sqrt3-\dfrac{2^{2x}}2$$
$$2^{2x}\left(1+\dfrac12\right)=3^x\left(\sqrt3+\dfrac1{\sqrt3}\right)$$
$$2^{2x}\cdot\dfrac32=3^x\dfrac4{\sqrt3}$$
Take log in both sides to get $$2x\ln2+\ln3-\ln2=x\ln3+2\ln2-\dfrac12\ln3$$
$$\ln2(2x-1-2)=\ln3(x-\dfrac12-1)$$
$$\iff(2\ln2-\ln3)(2x-3)=0$$
Can you take it from here?
lab bhattacharjee
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Sorry the question is -3^x/√3 on lhs side and 3^x into √3 on rhs side – tapadia newlon Jun 26 '15 at 16:18
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