Hello everyone,
I'm having trouble solving this equation and we are using logarithms to solve it.
Solve the exponential equation algebraically. Approximate the result to three decimal places.
Problem: \(\displaystyle 2^{3-x}=565\)
My Work:
\(\displaystyle 2^{3-x}=565\)
\(\displaystyle log2^{3-x}=log565\)
\(\displaystyle (3-x)log2=log656\)
\(\displaystyle 3-log2-log565=x\) I brought x one one side of the equation and everything else on the other side.
\(\displaystyle x=3-(\dfrac{log2}{log565})\) I used the quotient rule to turn subtraction into division, also brought x to the left side for convience.
\(\displaystyle x=3-(\dfrac{0.301}{2.752})\) I used the calculator and found values for log 2 and log 565.
\(\displaystyle x=2.89\)
This is the answer in my textbook:
\(\displaystyle 3-\dfrac{ln565}{ln2}\approx-6.142\)
I don't think it matters if you use natural logs (\(\displaystyle ln\)) or logs (\(\displaystyle log\))
I'm having trouble solving this equation and we are using logarithms to solve it.
Solve the exponential equation algebraically. Approximate the result to three decimal places.
Problem: \(\displaystyle 2^{3-x}=565\)
My Work:
\(\displaystyle 2^{3-x}=565\)
\(\displaystyle log2^{3-x}=log565\)
\(\displaystyle (3-x)log2=log656\)
\(\displaystyle 3-log2-log565=x\) I brought x one one side of the equation and everything else on the other side.
\(\displaystyle x=3-(\dfrac{log2}{log565})\) I used the quotient rule to turn subtraction into division, also brought x to the left side for convience.
\(\displaystyle x=3-(\dfrac{0.301}{2.752})\) I used the calculator and found values for log 2 and log 565.
\(\displaystyle x=2.89\)
This is the answer in my textbook:
\(\displaystyle 3-\dfrac{ln565}{ln2}\approx-6.142\)
I don't think it matters if you use natural logs (\(\displaystyle ln\)) or logs (\(\displaystyle log\))
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