<img src="https://tex.z-dn.net/?f=log%28x%2B3%29%20%3D%20log%208%20-%20log%202" id="TexFormula1" title="log(x+3) = log 8

zysi [14]

2 years ago

" alt="log(x+3) = log 8 - log 2" align="absmiddle" class="latex-formula">
agree or disagree and justify

Mathematics

1 answer:

choli [55]2 years ago

7 0

Answer:

Step-by-step explanation:

log (x+3)=log8-log2

log(x+3)=log(8/2)=log4

x+3=4

x=4-3=1

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2x + 10 + 3x = 7x + 12

wlad13 [49]

Answer:

Step-by-step explanation:

$2x+10+3x=7x+12\qquad\text{combine like terms}\\\\(2x+3x)+10=7x+12\\\\5x+10=7x+12\qquad\text{subtract 10 from both sides}\\\\5x+10-10=7x+12-10\\\\5x=7x+2\qquad\text{subtract}\ 7x\ \text{from both sides}\\\\5x-7x=7x-7x+2\\\\-2x=2\qquad\text{divide both sides by (-2)}\\\\\dfrac{-2x}{-2}=\dfrac{2}{-2}\\\\x=-1$

8 0

3 years ago

Read 2 more answers

Derivative of <br> <img src="https://tex.z-dn.net/?f=x%5E%7B3%7D" id="TexFormula1" title="x^{3}" alt="x^{3}" align="absmiddle" c

ch4aika [34]

$f(x)=x^3e^{2x}\\\\f'(x)=(x^3e^{2x})'\qquad\text{use}\ (f\cdot g)'(x)=f'(x)g(x)+f(x)g'(x)\\\\=(x^3)'(e^{2x})+(x^3)(e^{2x})'\\\\\text{use}\ (x^n)'=nx^{n-1}\ \text{and}\ (e^x)'=e^x\ \text{and}\ f'[g(x)]=f'(g(x))\cdot g'(x)\\\\=3x^2e^{2x}+x^3e^{2x}\cdot2\\\\=\boxed{3x^2e^{2x}+2x^3e^{2x}}$