(ii) Log2)(100x)+log2)(10x)=14+log((1)/(x))

2 min read Jun 11, 2024

Solving the Equation

In this article, we will solve the equation:

$\log_2(100x)+\log_2(10x)=14+\log_2\left(\frac{1}{x}\right)$

Using the property of logarithms, we can rewrite the left-hand side as:

$\log_2(100x)+\log_2(10x)=\log_2(1000x^2)$

Using the property of logarithms, we can rewrite the right-hand side as:

$14+\log_2\left(\frac{1}{x}\right)=14-\log_2(x)$

Equating the two expressions, we get:

$\log_2(1000x^2)=14-\log_2(x)$

To solve for x, we can rewrite the equation as:

$\log_2(x^2)=14-\log_2(1000)$

Using the property of logarithms, we can rewrite the equation as:

$\log_2(x^2)=\log_2\left(\frac{2^{14}}{1000}\right)$

Taking the exponential of both sides, we get:

$x^2=\frac{2^{14}}{1000}$

Solving for x, we get:

$x=\sqrt{\frac{2^{14}}{1000}}=\frac{2^7}{10}$

Therefore, the solution to the equation is:

$x=\frac{128}{10}=\boxed{12.8}$