mathematics

Question

Im not able to derive the value of X.plz someone derive the value of X.

Answer 1

$\begin{align*} &\text{assume } 2^{x} = a \\ &\Rightarrow y = 2a+a^{2}\\ &\Rightarrow a^{2}+2a-y=0 \\ &\Rightarrow a=\frac{-2\pm \sqrt{2^2-4.1.(-y)}}{2.1} \\ &\Rightarrow a=\sqrt{1+y} -1 \ \ \ \ \left [ \text{ x is real } \rightarrow \text{ a is real } \right ] \\ &\Rightarrow 2^{x}=\sqrt{1+y} -1 \\ &\Rightarrow x=\log_2(\sqrt{1+y} -1) \\ &\Rightarrow f^{-1}(x)=\log_2(\sqrt{1+x} -1) = \frac{\log_{10}(\sqrt{1+x} -1)}{\log_{10} 2} \\ \end{align*}$

Comments

but putting values is the best method.

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Ya it helps from the examination and time management perspective.

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But still it is not option A---option a is $[log(\sqrt{x+1})-1] not log(\sqrt{x+1}-1)$

Answer 2

The easiest way to answer this question is by putting values. In the function $\color{red}{y = 2*2^x  + 4^x}$, When $x = 1$, $y = 8$. So by substituing $x = 8$ in all options the one which gives $y = 1$ may be the answer.

Option(A) $\color{blue}{\frac{log(\sqrt{8 + 1} -1}{log(2)} = \frac{log2}{log2} = 1}$

Option(B) $\color{blue}{\frac{log3}{log2}}$

Option(C) $\color{blue}{log(\sqrt{8+1} -1) = log2}$

Option(D) $\color{blue}{log(\sqrt{8+1}) = log3}$

Hence, (A) is correct answer. If two options would have given answer as $1$, then check by substituting $x = 2$, but it becomes worse if there is an option (None of these). Then we need to solve it to be sure.

$\color{navy}{y = 2*2^x  + 4^x}$ ($\color{maroon}{4^x = 2^{2x}}$)

$\color{navy}{y = 2*2^x  + 2^{2x}}$

Adding $1$ to both sides.

$\color{navy}{y + 1= 2*2^x  + 2^{2x}+ 1}$

$\color{navy}{y + 1 = (2^x + 1)^2}$

$\color{navy}{ 2^x = \sqrt{y + 1}- 1}$

$\color{navy}{x = log_{2}(\sqrt{y + 1} -1)= \frac{log_{10}(\sqrt{y + 1} -1)}{log_{10}(2)}}$

Comments

Exactly!! Putting values and check is easy...