Question

drag the tiles to the correct boxes to complete the pairs.
match each equation with its solution.
\\(\ln (x + 5) = \ln (x - 1) + \ln (x + 1)\\) \\(e^{x^2} = e^{4x + 5}\\)
\\(\log_4 (5x^2 + 2) = \log_4 (x + 8)\\) \\(\log (x - 1) + \log 5x = 2\\)
\\(\left(-1, \frac{8}{5}\
ight)\\)
only 5
only 3
\\((-1, 5)\\)

Explanation:

Step1: Solve \(\ln(x + 5)=\ln(x - 1)+\ln(x + 1)\)

Use logarithm property \(\ln a+\ln b=\ln(ab)\), so right side is \(\ln[(x - 1)(x + 1)]=\ln(x^{2}-1)\). Then \(\ln(x + 5)=\ln(x^{2}-1)\), so \(x + 5=x^{2}-1\) (since \(\ln\) is one - to - one), \(x^{2}-x - 6 = 0\), factor: \((x - 3)(x+2)=0\), solutions \(x = 3\) or \(x=-2\). Check domain: \(x+5>0\), \(x - 1>0\), \(x + 1>0\). For \(x=-2\), \(x - 1=-3<0\) (invalid). For \(x = 3\), all conditions hold. So solution is \(x = 3\) (matches "only 3").

Step2: Solve \(e^{x^{2}}=e^{4x + 5}\)

Since \(e^{a}=e^{b}\) implies \(a = b\) (exponential function is one - to - one), so \(x^{2}=4x + 5\), \(x^{2}-4x - 5=0\), factor: \((x - 5)(x + 1)=0\), solutions \(x = 5\) or \(x=-1\). Check domain (exponential is defined for all real \(x\)). So solutions are \(x=-1\) and \(x = 5\) (matches \((-1,5)\)).

Step3: Solve \(\log_{4}(5x^{2}+2)=\log_{4}(x + 8)\)

Since \(\log_{4}a=\log_{4}b\) implies \(a = b\) (logarithm with base \(>1\) is one - to - one), so \(5x^{2}+2=x + 8\), \(5x^{2}-x - 6=0\), factor: \((5x - 6)(x + 1)=0\), solutions \(x=\frac{6}{5}\) or \(x=-1\). Check domain: \(5x^{2}+2>0\) (always true), \(x + 8>0\). For \(x=-1\), \(x + 8 = 7>0\); for \(x=\frac{6}{5}\), \(x + 8=\frac{6}{5}+8=\frac{46}{5}>0\). So solutions are \(x=-1\) and \(x=\frac{6}{5}\) (matches \((-1,\frac{6}{5})\)).

Step4: Solve \(\log(x - 1)+\log(5x)=2\)

Use \(\log a+\log b=\log(ab)\), so \(\log[5x(x - 1)] = 2\). Assume base 10, so \(5x(x - 1)=10^{2}=100\), \(5x^{2}-5x - 100 = 0\), divide by 5: \(x^{2}-x - 20=0\), factor: \((x - 5)(x + 4)=0\), solutions \(x = 5\) or \(x=-4\). Check domain: \(x - 1>0\) and \(5x>0\) (so \(x>1\)). For \(x=-4\), \(x - 1=-5<0\) (invalid). For \(x = 5\), all conditions hold. So solution is \(x = 5\) (matches "only 5").

Answer:

• \(\ln(x + 5)=\ln(x - 1)+\ln(x + 1)\) \(

ightarrow\) "only 3"

• \(e^{x^{2}}=e^{4x + 5}\) \(

ightarrow\) \((-1,5)\)

• \(\log_{4}(5x^{2}+2)=\log_{4}(x + 8)\) \(

ightarrow\) \((-1,\frac{6}{5})\)

• \(\log(x - 1)+\log(5x)=2\) \(

ightarrow\) "only 5"