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Drag the tiles to the correct boxes to complete the pairs.

Match each equation with its solution.

[tex]\[
\begin{array}{c}
\ln (x+5)=\ln (x-1)+\ln (x+1) \\
\log _4\left(5 x^2+2\right)=\log _4(x+8) \\
e^{x^2}=e^{4 x+5} \\
\log (x-1)+\log 5 x=2 \\
\end{array}
\][/tex]

[tex]\[
\begin{array}{c}
(-1, \frac{6}{5}) \longrightarrow \square \\
\text{only } 5 \rightarrow \square \\
\text{only } 3 \rightarrow \square \\
(-1, 5) \longrightarrow \square \\
\end{array}
\][/tex]

Sagot :

GinnyAnswer
Sure! Let's match each equation with its solution.

1. Given the equation [tex]\(\ln(x+5) = \ln(x-1) + \ln(x+1)\)[/tex], the solution is [tex]\( x = 3 \)[/tex].

2. Given the equation [tex]\(\log_4(5x^2+2) = \log_4(x+8)\)[/tex], the solutions are [tex]\( x = 1.2 \)[/tex] and [tex]\( x = -1.0 \)[/tex].

3. Given the equation [tex]\(e^{x^2} = e^{4x+5}\)[/tex], the solutions are [tex]\( x = 5.0 \)[/tex] and [tex]\( x = -1.0 \)[/tex].

4. Given the equation [tex]\(\log(x-1) + \log(5x) = 2\)[/tex], the solutions are [tex]\( x = 5.0 \)[/tex] and [tex]\( x = -4.0 \)[/tex].

Here are the matches:

- [tex]\(\ln(x+5) = \ln(x-1) + \ln(x+1)\)[/tex] only [tex]\(\rightarrow\)[/tex] only 3
- [tex]\(\log_4(5x^2+2) = \log_4(x+8)\)[/tex] [tex]\(\rightarrow\)[/tex] [tex]\((-1, 1.2)\)[/tex]
- [tex]\(e^{x^2} = e^{4x+5}\)[/tex] [tex]\(\rightarrow\)[/tex] [tex]\((-1, 5.0)\)[/tex]
- [tex]\(\log(x-1) + \log(5x) = 2\)[/tex] [tex]\(\rightarrow\)[/tex] [tex]\(5\)[/tex]

Hence:

- [tex]\(\ln(x+5) = \ln(x-1) + \ln(x+1)\)[/tex] [tex]\(\rightarrow\)[/tex] only 3
- [tex]\(\log_4(5x^2+2) = \log_4(x+8)\)[/tex] [tex]\(\rightarrow\)[/tex] [tex]\((-1, 1.2)\)[/tex]
- [tex]\(e^{x^2} = e^{4x+5}\)[/tex] [tex]\(\rightarrow\)[/tex] [tex]\((-1, 5.0)\)[/tex]
- [tex]\(\log(x-1) + \log(5x) = 2\)[/tex] [tex]\(\rightarrow\)[/tex] 5
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