# ln(2x-3)=ln 9-ln(x-3) Solve for x.

## Question:

{eq}\ln \left( {2x - 3} \right) = \ln 9 - \ln \left( {x - 3} \right) {/eq}

Solve for x.

## Logarithmic Function

The general notation for logarithmic function is {eq}f\left( x \right) = {\log _a}x {/eq}, where the dependent variable is {eq}f\left( x \right) {/eq}, independent variable is {eq}x {/eq} and the base of the function is {eq}a {/eq}.

Here, the value of the independent variable and the base should be greater than zero.

## Answer and Explanation:

**Given Data**

- The function is: {eq}\ln \left( {2x - 3} \right) = \ln 9 - \ln \left( {x - 3} \right) {/eq}

Solve the function using basic rule of logarithmic.

{eq}\begin{align*} \ln \left( {2x - 3} \right) &= \ln 9 - \ln \left( {x - 3} \right)\\ \ln \left( {2x - 3} \right) &= \ln \dfrac{9}{{\left( {x - 3} \right)}} \end{align*} {/eq}

Take antilog on both sides of the equation,

{eq}\begin{align*} \left( {2x - 3} \right) &= \dfrac{9}{{\left( {x - 3} \right)}}\\ \left( {2x - 3} \right)\left( {x - 3} \right) &= 9\\ 2{x^2} - 6x - 3x + 9 &= 9\\ 2{x^2} - 9x &= 0\\ x &= 0,4.5 \end{align*} {/eq}

Here, the value of {eq}x {/eq} can not be zero because logarithmic function will not be defined at {eq}x=0 {/eq}, so the value of {eq}x {/eq} is {eq}4.5 {/eq}.

Thus, the value of {eq}x {/eq} is {eq}4.5 {/eq}.

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