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Extraneous Solutions Calculator

Get detailed solutions to your math problems with our Extraneous Solutions step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here.

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Example

Solved Problems

Difficult Problems

1

Here, we show you a step-by-step solved example of extraneous solutions. This solution was automatically generated by our smart calculator:

$\log_2\left(x\right)+\log_2\left(x-3\right)=2$
2

Express the numbers in the equation as logarithms of base $2$

$\log_{2}\left(x\right)+\log_{2}\left(x-3\right)=\log_{2}\left(2^{2}\right)$
3

The sum of two logarithms of the same base is equal to the logarithm of the product of the arguments

$\log_{2}\left(x\left(x-3\right)\right)=\log_{2}\left(2^{2}\right)$
4

Use the following rule for logarithms: $\log_b(b^k)=k$

$\log_{2}\left(x\left(x-3\right)\right)=2$

Multiplying polynomials $x$ and $x-3$

$\log_{2}\left(x\cdot x-3x\right)=2$

When multiplying two powers that have the same base ($x$), you can add the exponents

$\log_{2}\left(x^2-3x\right)=2$
5

Multiplying polynomials $x$ and $x-3$

$\log_{2}\left(x^2-3x\right)=2$
6

Rewrite the number $2$ as a logarithm of base $2$

$\log_{2}\left(x^2-3x\right)=\log_{2}\left(2^{2}\right)$
7

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 $b$

$x^2-3x=2^{2}$
8

Calculate the power $2^{2}$

$x^2-3x=4$
9

Move everything to the left hand side of the equation

$x^2-3x-4=0$
10

Factor the trinomial $x^2-3x-4$ finding two numbers that multiply to form $-4$ and added form $-3$

$\begin{matrix}\left(1\right)\left(-4\right)=-4\\ \left(1\right)+\left(-4\right)=-3\end{matrix}$
11

Rewrite the polynomial as the product of two binomials consisting of the sum of the variable and the found values

$\left(x+1\right)\left(x-4\right)=0$
12

Break the equation in $2$ factors and set each equal to zero, to obtain

$x+1=0,\:x-4=0$
13

Solve the equation ($1$)

$x+1=0$
14

We need to isolate the dependent variable $x$, we can do that by simultaneously subtracting $1$ from both sides of the equation

$x+1-1=0-1$
15

Canceling terms on both sides

$x=-1$
16

Solve the equation ($2$)

$x-4=0$
17

We need to isolate the dependent variable $x$, we can do that by simultaneously subtracting $-4$ from both sides of the equation

$x-4+4=0+4$
18

Canceling terms on both sides

$x=4$
19

Combining all solutions, the $2$ solutions of the equation are

$x=-1,\:x=4$

Verify that the solutions obtained are valid in the initial equation

20

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

$x=4$

Final answer to the problem

$x=4$

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