Question

Given exponential equations,
a) \(3^{(4x\ -1) }=11\)
b) \(6^{x\ +\ 3}=3^{x}\)
To solve for x

Answer 1

Sub part a)
We have,
\(3^{(ax\ -\ 1)}=11\)
Applying log to the base 10 on both sides,
\(\Rightarrow\ \log_{10}\ 3^{(4x\ -\ 1)}=\ \log_{10}11\)
\(\Rightarrow\ (4x\ -\ 1)\log_{10}3=\log_{10}11\)
\(\Rightarrow\ 4x\ -\ 1\ = (\log_{10}11)/(\log_{10}3)\)
\(\Rightarrow\ 4x\ -\ 1\ =\ \log_{3}11\)
\(\Rightarrow\ x= (1/4)[(\log_{3}11)\ +\ 1]\)
\(\Rightarrow\ x = 0.795664585\)
(Rounded to 9 decimals)
Identities used:
\((\log_{a} x)/(\log_{a} y) = \log_{y} x\)
\(\log\ a^{b} = b\ \log\ a\)
Sub part b)
We have,
\(6^{x\ +\ 3} = 3^{x}\)
Applying log to base 2 on both sides,
\(\Rightarrow\ (x\ +\ 3) \log_{2}\ 6 = x(\log_{2} 3)\)
\(\Rightarrow\ (x\ +\ 3) \log_{2}\ (3 \times\ 2) = x (\log_{2} 3)\)
\(\Rightarrow\ (x\ +\ 3) [(\log_{2} 3)\ +\ (\log_{2} 2)] = x (\log_{2} 3)\)
\(\Rightarrow\ (x\ +\ 3) [(\log_{2} 3)\ +\ 1] = x(\log_{2} 3)\)
\(\Rightarrow\ x (\log_{2} 3)\ +\ 3 (\log_{2} 3)\ +\ x\ +\ 3 = x (\log_{2} 3)\)
\(\Rightarrow\ \log_{2} 3^{3}\ +\ x\ +\ 3 = 0\)
\(\Rightarrow\ \log_{2} 9\ +\ x\ +\ 3 = 0\)
\(\Rightarrow\ x =\ -3\ -\ \log_{2} 9\)
\(\Rightarrow\ x =\ -6.169925001\)
(Rounded to 9 decimals)
Identities used:
\(\log_{a} xy = (\log_{a} x)\ +\ (\log_{a} y)\)
\(\log a^{b} = b\ \log\ a\)