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log(x+1)=log x+1

$28$ . $\log (x+1)=\log x+1$

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Convert between exponential and logarithmic form

Full solution

Q.

28

.

\log (x+1)=\log x+1

Understand Domain: We need to solve the equation $\log(x+1) = \log x + 1$ . First, let's understand that the logarithm function is only defined for positive arguments. Therefore, both $x+1$ and $x$ must be greater than $0$ . $x+1 > 0$ implies $x > -1$ and $x > 0$ implies $x > 0$ . So, the domain for $x$ is $x > 0$ .
Set Logarithms Equal: Now, let's use the property of logarithms that states if $\log(a) = \log(b)$ , then $a = b$ . So, we can set the arguments of the logarithms equal to each other: $x+1 = x + 1$ .
Simplify Equation: This equation simplifies to $x + 1 = x + 1$ . Subtract $x$ from both sides to isolate the constant terms: $1 = 1$ .
Check Identity: We see that $1 = 1$ is a true statement, which means that the original equation is an identity for all $x$ in the domain $(x > 0)$ . Therefore, there is no unique solution for $x$ ; any positive $x$ satisfies the equation.

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