Here, we show you a step-by-step solved example of extraneous solutions. This solution was automatically generated by our smart calculator:
Express the numbers in the equation as logarithms of base $2$
The sum of two logarithms of the same base is equal to the logarithm of the product of the arguments
Use the following rule for logarithms: $\log_b(b^k)=k$
Multiplying polynomials $x$ and $x-3$
When multiplying two powers that have the same base ($x$), you can add the exponents
Multiplying polynomials $x$ and $x-3$
Rewrite the number $2$ as a logarithm of base $2$
For two logarithms of the same base to be equal, their arguments must be equal. In other words, if $\log(a)=\log(b)$ then $a$ must equal $b$
Calculate the power $2^{2}$
Move everything to the left hand side of the equation
Factor the trinomial $x^2-3x-4$ finding two numbers that multiply to form $-4$ and added form $-3$
Rewrite the polynomial as the product of two binomials consisting of the sum of the variable and the found values
Break the equation in $2$ factors and set each equal to zero, to obtain
Solve the equation ($1$)
We need to isolate the dependent variable $x$, we can do that by simultaneously subtracting $1$ from both sides of the equation
Canceling terms on both sides
Solve the equation ($2$)
We need to isolate the dependent variable $x$, we can do that by simultaneously subtracting $-4$ from both sides of the equation
Canceling terms on both sides
Combining all solutions, the $2$ solutions of the equation are
Verify that the solutions obtained are valid in the initial equation
The valid solutions to the logarithmic equation are the ones that, when replaced in the original equation, don't result in any logarithm of negative numbers or zero, since in those cases the logarithm does not exist
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