Resources

Resources

Bytelearn - cat image with glasses

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

solve for $x$ if $\log_{10}(3x+1) + \log_{10}(2x-5)=0$

View step-by-step help

View step-by-step help

Solve advanced linear inequalities

Full solution

Q. solve for

x

if

\log_{10}(3x+1) + \log_{10}(2x-5)=0

Combine Logarithmic Expressions: Combine the logarithmic expressions using the product rule for logarithms, which states that $\log(a) + \log(b) = \log(ab)$ . $\log_{10}(3x+1) + \log_{10}(2x-5) = 0$ $\log_{10}((3x+1)(2x-5)) = 0$
Convert to Exponential Form: Convert the logarithmic equation to its exponential form. Since $\log_{10}(y) = x$ is equivalent to $10^x = y$ , we can write: $\newline$ $10^0 = (3x+1)(2x-5)$
Simplify Exponential Expression: Simplify the exponential expression, knowing that $10^0 = 1$ . $\newline$ $1 = (3x+1)(2x-5)$
Expand Equation: Expand the right side of the equation. $\newline$ $1 = 6x^2 - 15x + 2x - 5$ $\newline$ $1 = 6x^2 - 13x - 5$
Set Quadratic Equation to Zero: Move all terms to one side to set the quadratic equation to zero. $\newline$ $0 = 6x^2 - 13x - 5 - 1$ $\newline$ $0 = 6x^2 - 13x - 6$
Factor or Use Quadratic Formula: Factor the quadratic equation, if possible, or use the quadratic formula to find the values of $x$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ . In this case, $a = 6$ , $b = -13$ , and $c = -6$ .
Discriminant: $\Delta = b^2 - 4ac = (-13)^2 - 4(6)(-6) = 169 + 144 = 313$
Since the discriminant is positive, there are two real solutions.
Calculate Solutions: Calculate the solutions using the quadratic formula. $\newline$ $x = \frac{-(-13) \pm \sqrt{313}}{2 \cdot 6}$ $\newline$ $x = \frac{13 \pm \sqrt{313}}{12}$
Check Validity of Solutions: Check for extraneous solutions by plugging the values of $x$ back into the original logarithmic expressions to ensure that the arguments of the logarithms are positive. $\newline$ For $x = \frac{13 + \sqrt{313}}{12}$ : $\newline$ $3x+1 = 3\left(\frac{13 + \sqrt{313}}{12}\right) + 1 > 0$ $\newline$ $2x-5 = 2\left(\frac{13 + \sqrt{313}}{12}\right) - 5 > 0$ $\newline$ Both expressions are positive, so this is a valid solution.
Check Second Solution: Check the second solution for $x = \frac{13 - \sqrt{313}}{12}$ : $3x+1 = 3\left(\frac{13 - \sqrt{313}}{12}\right) + 1$ $2x-5 = 2\left(\frac{13 - \sqrt{313}}{12}\right) - 5$ We need to check if these expressions are positive.
Check Second Solution: Check the second solution for $x = \frac{13 - \sqrt{313}}{12}$ :
$3x+1 = 3\left(\frac{13 - \sqrt{313}}{12}\right) + 1$
$2x-5 = 2\left(\frac{13 - \sqrt{313}}{12}\right) - 5$
We need to check if these expressions are positive.Calculate the expressions to ensure they are positive.
$3x+1 = 3\left(\frac{13 - \sqrt{313}}{12}\right) + 1$
This expression is positive because $\sqrt{313}$ is less than $13$ , so $(13 - \sqrt{313})$ is positive, and multiplying by $3$ and adding $1$ keeps it positive.
$2x-5 = 2\left(\frac{13 - \sqrt{313}}{12}\right) - 5$
This expression is negative because $\sqrt{313}$ is greater than $3x+1 = 3\left(\frac{13 - \sqrt{313}}{12}\right) + 1$ $1$ , so $(13 - \sqrt{313})$ is negative, and multiplying by $3x+1 = 3\left(\frac{13 - \sqrt{313}}{12}\right) + 1$ $3$ and subtracting $3x+1 = 3\left(\frac{13 - \sqrt{313}}{12}\right) + 1$ $4$ keeps it negative.
Since $3x+1 = 3\left(\frac{13 - \sqrt{313}}{12}\right) + 1$ $5$ is negative, this solution is not valid for the logarithmic equation.

More problems from Solve advanced linear inequalities

Question

-2 x+b y=59

\newline

5 x-y=1

\newline

If the system of equations has exactly one solution

(x, y)

x=2

, what is the value of

b

Posted 2 months ago

Question

\newline

\frac{\sqrt[3]{2}}{\sqrt[3]{-128}}=

Posted 18 days ago

Question

\newline

\frac{\sqrt[4]{2}}{\sqrt[4]{162}}=

Posted 4 months ago

Question

\newline

\sqrt[5]{-16} \cdot 2^{\frac{1}{5}}=

Posted 3 months ago

Question

y

\newline

\begin{array}{l} \frac{y-1}{3 y+7}=\frac{1}{10} \\ y=\square \end{array}

Posted 3 months ago

Question

x

\newline

Your answer must be simplified.

\newline

-30<\frac{x}{-4}

Posted 2 months ago

Question

x

\newline

Your answer must be simplified.

\newline

-36+x<8

Posted 2 months ago

Question

x

\newline

Your answer must be simplified.

\newline

2 x<15

Posted 3 months ago

Question

x

\newline

Your answer must be simplified.

\newline

x-6 \leq 3

Posted 2 months ago

Question

x

\newline

Your answer must be simplified.

\newline

7>\frac{x}{4}

Posted 3 months ago