Question

How do you solve ${{4}^{x}}-{{2}^{x+1}}-15=0$?

Answer 1

Any equation can be solved by taking all the constants to one side and all the unknowns to the other side of the equation. The constant side must be solved step-by-step to get through the solution. We can use the distributive property and do the addition, subtraction, multiplication and division operations wherever necessary in such a way to simplify the equation.

Complete step by step answer:
As per the given question, we are provided with an equation which is to be simplified to get the solution of the equation. A solution is that which when substituted back into the equation, both the sides of the equation will be equal. Here, the given equation is
\Rightarrow $$$${{4}^{x}}-{{2}^{x+1}}-15=0
Let us assume ${{2}^{x}}=a$ then the equation becomes
\Rightarrow $$$${{a}^{2}}-2a-15=0
Now we add 16 on both sides. We get

& \Rightarrow {{a}^{2}}-2a-15+16=0+16 \\\ & \Rightarrow {{a}^{2}}-2a+1=16 \\\ \end{aligned}$$ We know that $${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab-{{b}^{2}}$$. From this we can rewrite the equation as $$\begin{aligned} & \Rightarrow {{\left( a-1 \right)}^{2}}=16 \\\ & \Rightarrow {{\left( a-1 \right)}^{2}}={{4}^{2}} \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow a-1=4,a-1=-4 \\\ & \Rightarrow a=5,-3 \\\ \end{aligned}$$ Now we substitute the value of a then the values become $$\Rightarrow $$$${{2}^{x}}=5,-3$$ Since $${{2}^{x}}$$ cannot be negative then $${{2}^{x}}=5$$. Now we apply logarithm with base 10 on both sides. $$\Rightarrow $$$${{\log }_{10}}{{2}^{x}}={{\log }_{10}}5$$ According to properties of logarithm, $${{\log }_{10}}{{a}^{n}}=n{{\log }_{10}}a$$. We can rewrite the above expression as $$\Rightarrow $$$$x{{\log }_{10}}2={{\log }_{10}}5$$ On dividing the expression with $${{\log }_{10}}2$$ on both sides. We get $$\begin{aligned} & \Rightarrow \dfrac{x{{\log }_{10}}2}{{{\log }_{10}}2}=\dfrac{{{\log }_{10}}5}{{{\log }_{10}}2} \\\ & \Rightarrow x=\dfrac{{{\log }_{10}}5}{{{\log }_{10}}2} \\\ \end{aligned}$$ The value of $${{\log }_{10}}5$$ is $$0.69897000433$$ and the value of $${{\log }_{10}}2$$ is $$0.30102999566$$. Then the value of x will be $$2.32192809$$. It can be rounded off to 3 decimals. Then x will be $$2.322$$. **Note:** In order to solve such types of questions, we need to have enough knowledge over logarithms, exponents, quadratic equations and their properties. We also need to know the algebraic formulae to simplify the expressions. We must avoid calculation mistakes to get the expected answers.