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

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

2020-11-18
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$$

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