Question

The exponential equation and approximate the result, correct to 9 decimal places. a) 3^{(4x-1)}=11 b) 6^{x+3}=3^{x} To solve for x.

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asked 2021-05-26
The exponential equation and approximate the result, correct to 9 decimal places.
a) \(3^{(4x-1)}=11\)
b) \(6^{x+3}=3^{x}\)
To solve for x.

Expert Answers (1)

2021-05-27
Step 1
a) We have,
\(3^{(4x-1)}=11\)
Applying log to the base 10 on both sides,
\(\Rightarrow\log_{10}3^{(4x-1)}=\log_{10}11\)
\(\Rightarrow\ (4x-3)\log_{10}3=\log_{10}11\)
\(\Rightarrow4x-1=(log_{10}11)/(\log_{10}3)\)
\(\Rightarrow4x-1=\log_{3}11\)
\(\Rightarrow\ x=(1/4)[(\log_{3}11)+1]\)
\(\Rightarrow\ x=0.795664585\)
(Rounded to 9 decimals)
Identities used :
\((\log_{a}x)/(\log_{a}y)=\log_{y}x\)
\(\log\ a^{b}=b\log\ a\)
Step 2
b) We have,
\(6^{x+3}=3^{x}\)
Applying log to base 2 on both sides,
\(\Rightarrow(x+3)\log_{2}6=x(\log_{2}3)\)
\(\Rightarrow(x+3)\log_{2}(3\times2)=x(\log_{2}3)\)
\(\Rightarrow(x+3)[(\log_{2}3)+(\log_{2})]=x(\log_{2}3)\)
\(\Rightarrow(x+3)[(\log_{2}3)+1]=x(\log_{2}3)\)
\(\Rightarrow\ x(\log_{2}3)+3(\log_{2}3)+x+3=x(\log_{2}3)\)
\(\Rightarrow\log_{2}3^{3}+x+3=0\)
\(\Rightarrow\log_{2}9+x+3=0\)
\(\Rightarrow\ x=-3-\log_{2}9\)
\(\Rightarrow\ x=-6.169925001\)
(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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