Answer:
8+6-10+2+15-3-7-5+2-15-8+10 = 5
Step-by-step explanation:
<span>ln(x+2)-ln(4x+3)=ln(1/2*x)
Using properties of logarithms
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![\frac{ln(x+2)}{ln(4x+3)} = ln \frac{x}{2} \\ \\ \frac{x+2}{4x+3} = \frac{x}{2} \\ \\2(x+2)=x(4x+3) 2x+4=4x^2+3x \\ \\ 4 x^{2} +x-4=0 \\ \\ x= \frac{-b+/- \sqrt{b^{2}-4ac} }{2a} \\ \\ x= \frac{ -1+/-\sqrt{1+64} }{8} \\ \\ x_{1} = \frac{-1+ \sqrt{65} }{8} \\ \\ x_{2} = \frac{-1- \sqrt{65} }{8} Check: When you substitute x_{2} into \\ \\ ln(4x+3)=ln(4* \frac{-1- \sqrt{65} }{8} ) =ln( \frac{-1- \sqrt{65} }{2} ) you will get negative number under ln, that is impossible , ](https://tex.z-dn.net/?f=%20%5Cfrac%7Bln%28x%2B2%29%7D%7Bln%284x%2B3%29%7D%20%3D%20ln%20%20%5Cfrac%7Bx%7D%7B2%7D%0A%5C%5C%20%5C%5C%20%20%5Cfrac%7Bx%2B2%7D%7B4x%2B3%7D%20%3D%20%5Cfrac%7Bx%7D%7B2%7D%20%0A%5C%5C%20%5C%5C2%28x%2B2%29%3Dx%284x%2B3%29%0A%0A2x%2B4%3D4x%5E2%2B3x%0A%5C%5C%20%5C%5C%204%20x%5E%7B2%7D%20%2Bx-4%3D0%0A%5C%5C%20%5C%5C%20x%3D%20%5Cfrac%7B-b%2B%2F-%20%5Csqrt%7Bb%5E%7B2%7D-4ac%7D%20%7D%7B2a%7D%20%0A%5C%5C%20%5C%5C%20x%3D%20%5Cfrac%7B%20-1%2B%2F-%5Csqrt%7B1%2B64%7D%20%7D%7B8%7D%20%0A%5C%5C%20%5C%5C%20x_%7B1%7D%20%3D%20%20%5Cfrac%7B-1%2B%20%5Csqrt%7B65%7D%20%7D%7B8%7D%20%0A%5C%5C%20%5C%5C%20x_%7B2%7D%20%3D%20%5Cfrac%7B-1-%20%5Csqrt%7B65%7D%20%7D%7B8%7D%20%0A%0ACheck%3A%0A%0AWhen%20you%20substitute%20x_%7B2%7D%20into%0A%5C%5C%20%5C%5C%20ln%284x%2B3%29%3Dln%284%2A%20%5Cfrac%7B-1-%20%5Csqrt%7B65%7D%20%7D%7B8%7D%20%29%20%3Dln%28%20%5Cfrac%7B-1-%20%5Csqrt%7B65%7D%20%7D%7B2%7D%20%29%20%0A%0Ayou%20will%20get%20negative%20number%20under%20ln%2C%20%20that%20is%20impossible%20%2C%0A%0A%0A)
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so x2 is not a solution of this logarithmic equation.
Only x1 is a solution.
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I suck at fractiona but u start off with 2(5)+8=1/4x U get 10+8=1/4x Then 18=1/4x from there im lost
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