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How does one solve $\log (2x - 1) = 3$ ?

seo-qna
Last updated date: 22nd Feb 2024
Total views: 337.5k
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IVSAT 2024
Answer
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Hint:For simplifying the original equation , take base ten exponential of both sides of the equation, then apply the logarithm formula ${b^{{{\log }_b}a}} = a$ to simplify the equation. Then recall that the logarithm function says $\log x$ is only defined when $x$ is greater than zero.

Formula used:
We used logarithm property i.e.
${b^{{{\log }_b}a}} = a$ , the logarithm function says $\log x$ is only defined when $x$ is greater than zero.

Complete solution step by step:
It is given that ,
$\log (2x - 1) = 3$
We have to solve for $x$ ,
Now , by assuming the base of the logarithm to be $10$ ,then take the base $10$ exponential of both sides of the equation, we will get the following result ,
${10^{{{\log }_{10}}(2x - 1)}} = {10^3}$
By applying the logarithm formula ${b^{{{\log }_b}a}} = a$ . we will get the following result ,
$2x - 1 = 1000$ ,
Now simplify the above equation as ,
$
\Rightarrow 2x = 1000 + 1 \\
\Rightarrow 2x = 1001 \\
\Rightarrow x = \dfrac{{1001}}{2} \\
$
Now recall that the logarithm function says $\log x$ is only defined when $x$ is greater than zero.
Therefore, in our original equation $\log (2x - 1) = 3$ ,
Here we must have ,
$(2x - 1) > 0$ ,
We have $x = \dfrac{{1001}}{2}$
We get ,
$1000 > 0$
Therefore, we have our solution i.e., $x = \dfrac{{1001}}{2}$ .

Note: The logarithm function says $\log x$ is only defined when $x$ is greater than zero. While defining logarithm function one should remember that the base of the log must be a positive real number and not equals to one . At the end we must recall that the logarithm function says $\log x$ is only defined when $x$ is greater than zero. While performing logarithm properties we have remember certain conditions , our end result must satisfy domain of that logarithm