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How to solve $\log \left( -2x+9 \right)=\log \left( 7-4x \right)$?

Answer
VerifiedVerified
542.1k+ views
Hint: In this question we have been given with logarithmic functions on either side of the $=$ sign therefore, we will first remove the log and equate the terms which are present in the parenthesis. Sine the given equation will be a linear equation, we will rearrange the terms in the equation to get the value of $x$ which is the final answer.

Complete step by step solution:
We have the expression given to us as:
$\Rightarrow \log \left( -2x+9 \right)=\log \left( 7-4x \right)$
Since the similar log is present in both sides of the expression, the terms in the parenthesis of each term are equal to each other therefore, we can write the expression as:
$\Rightarrow -2x+9=7-4x$
Now we will take the similar terms on the same side, on transferring $-4x$ from the right-hand side to the left-hand side and transferring $9$ from the left-hand side to the right-hand side, we get:
$\Rightarrow -2x+4x=7-9$
On simplifying the left-hand side, we get:
$\Rightarrow 2x=7-9$
On simplifying the right-hand side, we get:
$\Rightarrow 2x=-2$
On dividing both the sides of the expression by $2$, we get:
$\Rightarrow x=-1$, which is the final answer.

Note: It is to be noted that the value inside the parenthesis cannot be negative as log of negative numbers does not exist. To verify, we will substitute the value of $x=-1$ in the left-hand side and the right-hand side of the expression to verify.
On substituting $x=-1$ in the left-hand side of the equation, we get:
$\Rightarrow \log \left( -2\left( -1 \right)+9 \right)$
On simplifying we get:
$\Rightarrow \log \left( 2+9 \right)$
On adding the terms, we get:
$\Rightarrow \log \left( 11 \right)$, since $11$ is a positive number therefore, the left-hand side is valid.
On substituting $x=-1$ in the right-hand side of the equation, we get:
$\Rightarrow \log \left( 7-4\left( -1 \right) \right)$
On simplifying we get:
$\Rightarrow \log \left( 7+4 \right)$
On adding the terms, we get:
$\Rightarrow \log \left( 11 \right)$, since $11$ is a positive number and is equal to the value on the left-hand side, the right-hand side is valid.
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