The main objective of this question is to use the one-to-one property of logarithms to conclude $\ln x=\ln y$.

A logarithm can be regarded as the number of powers to which a number must be raised to obtain some other values. It is one of the very suitable ways to illustrate large numbers. It is also known as the opposite of exponentiation. More generally, the logarithm of a given number $x$ is the exponent to which another fixed number, the base $a$, must be raised to produce $x$.

The logarithm to the base of constant $e$ is said to be the natural logarithm of a number where $e$ is approximately equal to $2.178$. For instance, consider an exponential function $e^x$ then $\ln(e^x)=e$. The natural logarithm contains the same properties as the common logarithm.

According to the one-to-one property of logarithmic functions, for any positive real numbers $x,y$ and $a\neq 1$, $\log_ax=\log_ay$ if and only if $x=y$.

And so, a similar property applies to the natural logarithm.

## Expert Answer

A function $f(x)$ is said to be one-to-one if $f(x_1)=f(x_2)\implies x_1=x_2$.

It is given that:

$\ln x=\ln y$

Applying exponentiation on both sides, we get:

$e^{\ln x}=e^{\ln y}$

$x=y$

So, by one-to-one property of natural logarithm:

If $\ln x=\ln y$ then $x=y$.

## Example 1

Solve $\ln(4x-3)-\ln(3)=\ln(x+1)$ by using the one-to-one property of natural logarithm.

### Solution

First, apply the quotient rule of logarithm as:

$\ln\left(\dfrac{4x-3}{3}\right)=\ln(x+1)$

Now, apply the one-to-one property of logarithm:

$e^{\ln\left(\dfrac{4x-3}{3}\right)}=e^{\ln(x+1)}$

$\dfrac{4x-3}{3}=x+1$

Multiply both sides of the above equation by $3$ to get:

$4x-3=3(x+1)$

$4x-3=3x+3$

Solve to obtain $x$ as:

$4x-3x=3+3$

$x=6$

## Example 2

Solve the following equation using the one-to-one property of the natural logarithm.

$\ln(x^2)=\ln(4x+5)$

### Solution

Applying the one-to-one property on given equation as:

$e^{\ln(x^2)}=e^{\ln(4x+5)}$

$x^2=4x+5$

$x^2-4x-5=0$

Factorize the above logarithmic equation as:

$x^2+x-5x-5=0$

$x(x+1)-5(x+1)=0$

$(x+1)(x-5)=0$

$x+1=0$ or $x-5=0$

$x=-1$ or $x=5$

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