Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

log3(72x)=5(4x+11)\log 3^{(7-2x)}=5^{(4x+11)}

Full solution

Q. log3(72x)=5(4x+11)\log 3^{(7-2x)}=5^{(4x+11)}
  1. Recognize Equation Involves Logarithms: First, we need to recognize that the equation involves logarithms and exponents. We will start by simplifying the logarithmic part of the equation.\newlinelog3(72x)=5(4x+11)\log 3^{(7-2x)} = 5^{(4x+11)}\newlineUsing the property of logarithms that logab=bloga\log a^b = b \cdot \log a, we can simplify the left side of the equation:\newline(72x)log3=5(4x+11)(7 - 2x) \cdot \log 3 = 5^{(4x+11)}
  2. Simplify Left Side of Equation: Next, we need to deal with the right side of the equation. Since there is no simple way to convert 54x+115^{4x+11} to a base of 33 or to a logarithmic form that would easily allow us to solve for xx, we will need to use a different strategy. We can apply logarithms to both sides of the equation to bring the exponents down. We will use the natural logarithm (ln\ln) for this purpose:\newlineln((72x)log3)=ln(54x+11)\ln((7 - 2x) \cdot \log 3) = \ln(5^{4x+11})
  3. Apply Natural Logarithm: We apply the natural logarithm to both sides of the equation to bring down the exponents:\newlineln(372x)=ln(54x+11)\ln(3^{7-2x}) = \ln(5^{4x+11})\newlineUsing the property of logarithms that ln(ab)=bln(a)\ln(a^b) = b \cdot \ln(a), we can simplify both sides:\newline(72x)ln(3)=(4x+11)ln(5)(7 - 2x) \cdot \ln(3) = (4x + 11) \cdot \ln(5)
  4. Simplify Linear Equation: Now we have a linear equation in terms of xx that we can solve. We will distribute ln(3)\ln(3) and ln(5)\ln(5) on both sides:\newline7×ln(3)2x×ln(3)=4x×ln(5)+11×ln(5)7 \times \ln(3) - 2x \times \ln(3) = 4x \times \ln(5) + 11 \times \ln(5)
  5. Collect Like Terms: Next, we will collect like terms and move all terms involving xx to one side of the equation and constants to the other side: 2xln(3)4xln(5)=11ln(5)7ln(3)-2x \cdot \ln(3) - 4x \cdot \ln(5) = 11 \cdot \ln(5) - 7 \cdot \ln(3)
  6. Factor Out xx: We can factor out xx from the left side of the equation:\newlinex(2ln(3)4ln(5))=11ln(5)7ln(3)x \cdot (-2 \cdot \ln(3) - 4 \cdot \ln(5)) = 11 \cdot \ln(5) - 7 \cdot \ln(3)
  7. Divide to Solve for x: Now we will divide both sides by (2×ln(3)4×ln(5))(-2 \times \ln(3) - 4 \times \ln(5)) to solve for x:\newlinex=11×ln(5)7×ln(3)2×ln(3)4×ln(5)x = \frac{11 \times \ln(5) - 7 \times \ln(3)}{-2 \times \ln(3) - 4 \times \ln(5)}
  8. Calculate x Value: Finally, we can calculate the value of xx using the values of ln(3)\ln(3) and ln(5)\ln(5):x=11ln(5)7ln(3)2ln(3)4ln(5)x = \frac{11 \cdot \ln(5) - 7 \cdot \ln(3)}{-2 \cdot \ln(3) - 4 \cdot \ln(5)}

More problems from Evaluate rational exponents