Question

question 9 of 10
what is the solution to the equation below? round your answer to two decimal places.
4 ( 8^x = 11.48 )

a. ( x approx 0.36 )
b. ( x approx 0.51 )
c. ( x approx 1.97 )
d. ( x approx 1.05 )

Explanation:

Step1: Take log on both sides

Take the logarithm of both sides of the equation \(8^{x}=11.48\). Using the property of logarithms \(\log(a^{b}) = b\log(a)\), we get \(x\log(8)=\log(11.48)\).

Step2: Solve for x

We can solve for \(x\) by dividing both sides by \(\log(8)\). So \(x = \frac{\log(11.48)}{\log(8)}\). Using a calculator, \(\log(11.48)\approx1.06\) and \(\log(8)\approx0.9031\). Then \(x=\frac{1.06}{0.9031}\approx1.17\)? Wait, no, let's recalculate. Wait, actually, using natural logarithm: \(\ln(11.48)\approx2.44\), \(\ln(8)\approx2.079\), so \(x=\frac{2.44}{2.079}\approx1.17\)? Wait, no, maybe I made a mistake. Wait, the options have 1.05. Wait, let's check again. Wait, \(8^{1.05}\): \(8^{1}=8\), \(8^{0.05}=e^{0.05\ln(8)}\approx e^{0.05\times2.079}\approx e^{0.10395}\approx1.11\), so \(8\times1.11 = 8.88\), no. Wait, \(8^{1.05}\) is not right. Wait, maybe I used wrong logs. Wait, let's use common logarithm (base 10). \(\log_{10}(11.48)\approx1.06\), \(\log_{10}(8)\approx0.9031\), so \(x = 1.06 / 0.9031\approx1.17\)? But the options have D as 1.05. Wait, maybe I miscalculated. Wait, let's compute \(8^{1.05}\): \(8^{1.05}=8^{1 + 0.05}=8\times8^{0.05}\). \(8^{0.05}=e^{0.05\ln8}\approx e^{0.05\times2.07944}\approx e^{0.10397}\approx1.11\), so \(8\times1.11 = 8.88\), no. Wait, \(8^{1.05}\) is 8.88, but 11.48 is larger. Wait, maybe I made a mistake in the equation. Wait, the equation is \(8^{x}=11.48\). Let's try x=1.05: \(8^{1.05}\). Let's calculate \(\ln(8^{1.05})=1.05\ln(8)\approx1.05\times2.0794 = 2.183\), and \(\ln(11.48)\approx2.44\), so 2.183 < 2.44. So x should be larger. Wait, the options: C is 1.97, D is 1.05, B is 0.51, A is 0.36. Wait, maybe I messed up. Wait, let's check \(8^{1.05}\): no, wait, maybe the equation is \(4\times8^{x}=11.48\)? Wait, the original problem says "4 \(8^{x}=11.48\)"? Wait, the user's image shows "4 \(8^{x}=11.48\)"? Wait, maybe it's \(4\times8^{x}=11.48\)? Oh! Maybe I misread the equation. If it's \(4\times8^{x}=11.48\), then \(8^{x}=11.48/4 = 2.87\). Then take log: \(x\log8=\log2.87\), so \(x=\log2.87/\log8\). \(\log2.87\approx0.4579\), \(\log8\approx0.9031\), so \(x\approx0.4579/0.9031\approx0.51\). Ah! That's option B. So I misread the equation: it's \(4\times8^{x}=11.48\), not \(8^{x}=11.48\). So step1: Divide both sides by 4: \(8^{x}=11.48/4 = 2.87\). Step2: Take log: \(x\log8=\log2.87\). Step3: Solve for x: \(x=\log2.87/\log8\approx0.4579/0.9031\approx0.51\).

Answer:

B. \(x \approx 0.51\)