Question

question 6/24
solve the following equation. round to 2 decimal places.
$3\cdot 2^{x + 1}-10 = 11$
$x = $

Explanation:

Step1: Isolate the exponential term

Add 10 to both sides of the equation \(3\cdot2^{x + 1}-10 = 11\).
We get \(3\cdot2^{x + 1}=11 + 10\), which simplifies to \(3\cdot2^{x + 1}=21\).

Step2: Solve for the exponential term

Divide both sides by 3: \(\frac{3\cdot2^{x + 1}}{3}=\frac{21}{3}\), so \(2^{x + 1}=7\).

Step3: Take the logarithm of both sides

Take the natural logarithm (ln) of both sides: \(\ln(2^{x + 1})=\ln(7)\).
Using the logarithm power rule \(\ln(a^b)=b\ln(a)\), we have \((x + 1)\ln(2)=\ln(7)\).

Step4: Solve for x

First, divide both sides by \(\ln(2)\): \(x + 1=\frac{\ln(7)}{\ln(2)}\).
Then subtract 1 from both sides: \(x=\frac{\ln(7)}{\ln(2)}-1\).

Step5: Calculate the value

We know that \(\ln(7)\approx1.9459\) and \(\ln(2)\approx0.6931\).
So \(\frac{\ln(7)}{\ln(2)}\approx\frac{1.9459}{0.6931}\approx2.8074\).
Then \(x\approx2.8074 - 1=1.8074\), and rounding to two decimal places gives \(x\approx1.81\).

Answer:

\(1.81\)