QUESTION IMAGE
Question
solve. round your answer to the nearest thousandth.
$7^{4x - 1} = 32$
$x = \square$
Step1: Take log of both sides
Take the natural logarithm (ln) of both sides of the equation \(7^{4x - 1}=32\). Using the property \(\ln(a^b)=b\ln(a)\), we get:
\(\ln(7^{4x - 1})=\ln(32)\)
\((4x - 1)\ln(7)=\ln(32)\)
Step2: Solve for \(4x - 1\)
Divide both sides by \(\ln(7)\) to isolate \(4x - 1\):
\(4x - 1=\frac{\ln(32)}{\ln(7)}\)
Step3: Solve for \(4x\)
Add 1 to both sides:
\(4x=\frac{\ln(32)}{\ln(7)} + 1\)
Step4: Solve for \(x\)
Divide both sides by 4:
\(x=\frac{\frac{\ln(32)}{\ln(7)} + 1}{4}\)
Now, calculate the numerical value. We know that \(\ln(32)\approx3.4657\) and \(\ln(7)\approx1.9459\).
First, calculate \(\frac{\ln(32)}{\ln(7)}\approx\frac{3.4657}{1.9459}\approx1.7809\).
Then, \(\frac{\ln(32)}{\ln(7)} + 1\approx1.7809 + 1 = 2.7809\).
Finally, \(x=\frac{2.7809}{4}\approx0.695\).
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\(0.695\)