Question

1 solve the following equation. round your answers to the nearest hundredths. $6 \cdot 3^{4x} = 60$ a 0.52 b 0.5239 c no solution d 0.53

Explanation:

Step1: Isolate the exponential term

Divide both sides of the equation \(6\cdot3^{4x}=60\) by 6:
\(3^{4x}=\frac{60}{6} = 10\)

Step2: Take logarithm on both sides

Take the natural logarithm (ln) of both sides:
\(\ln(3^{4x})=\ln(10)\)
Using the logarithm power rule \(\ln(a^b)=b\ln(a)\), we get:
\(4x\ln(3)=\ln(10)\)

Step3: Solve for x

First, divide both sides by \(4\ln(3)\):
\(x=\frac{\ln(10)}{4\ln(3)}\)
Calculate the values: \(\ln(10)\approx2.3026\), \(\ln(3)\approx1.0986\)
So \(x = \frac{2.3026}{4\times1.0986}=\frac{2.3026}{4.3944}\approx0.5239\)
Now, round to the nearest hundredths. The thousandths place is 3, which is less than 5, so we round down. So \(x\approx0.52\) (Wait, no, 0.5239 rounded to nearest hundredths: the number is 0.52 (tenths and hundredths) and the next digit is 3, so it stays 0.52? Wait, no, 0.5239: the hundredths digit is 2, the thousandths is 3. Wait, maybe I made a mistake. Wait, 0.5239: to nearest hundredths, look at the third decimal place (thousandths). 3 is less than 5, so we keep the hundredths digit as is. So 0.52? But option A is 0.52, option B is 0.5239, option D is 0.53. Wait, maybe I miscalculated. Wait, let's recalculate \(x=\frac{\ln(10)}{4\ln(3)}\). \(\ln(10)\approx2.302585093\), \(\ln(3)\approx1.098612289\). Then \(4\ln(3)=4\times1.098612289 = 4.394449156\). Then \(\frac{2.302585093}{4.394449156}\approx0.5239\). Now, rounding 0.5239 to the nearest hundredths: the hundredths place is 2, the thousandths is 3. Wait, no, 0.5239: the digits are 0 (units), 5 (tenths), 2 (hundredths), 3 (thousandths), 9 (ten - thousandths). To round to nearest hundredths, we look at the thousandths digit (3). Since 3 < 5, we round down, so 0.52. But wait, maybe the original equation is \(6\cdot3^{4x}=60\), let's check with x = 0.52: \(3^{4\times0.52}=3^{2.08}\). \(3^{2}=9\), \(3^{2.08}=3^{2 + 0.08}=9\times3^{0.08}\). \(3^{0.08}\approx e^{0.08\ln(3)}\approx e^{0.08\times1.0986}\approx e^{0.0879}\approx1.0919\). So \(9\times1.0919\approx9.827\). Then \(6\times9.827\approx58.96\), which is close to 60. If x = 0.5239: \(4x = 2.0956\), \(3^{2.0956}\). \(3^{2}=9\), \(3^{0.0956}\approx e^{0.0956\times1.0986}\approx e^{0.1051}\approx1.1103\). Then \(9\times1.1103 = 9.9927\), \(6\times9.9927\approx59.956\approx60\). Now, rounding 0.5239 to nearest hundredths: the number is 0.52 (hundredths place) and the next digit is 3, so we keep it 0.52? But option A is 0.52, option B is 0.5239 (unrounded), option D is 0.53. Wait, maybe the problem has a typo, or maybe I misread the exponent. Wait, the original equation is \(6\cdot3^{4x}=60\). Let's check the options. Option A is 0.52, option B is 0.5239, option D is 0.53. When we calculate 0.5239 rounded to nearest hundredths: look at the thousandths digit, which is 3, so we round down, so 0.52. But wait, 0.5239 is approximately 0.52 when rounded to the nearest hundredth? Wait, no: 0.5239, the first decimal is 5 (tenths), second is 2 (hundredths), third is 3 (thousandths), fourth is 9 (ten - thousandths). To round to the nearest hundredth, we look at the thousandths place (3). Since 3 < 5, we round down, so 0.52. But let's check the calculation again. \(x=\frac{\ln(10)}{4\ln(3)}\approx\frac{2.302585}{4\times1.098612}\approx\frac{2.302585}{4.394448}\approx0.5239\). Rounded to the nearest hundredth: 0.52 (because the third decimal is 3, which is less than 5). So the answer should be A. 0.52.

Answer:

A. 0.52