What is the true solution to 2 in e^in5x =2 in 15

Mathematics

2 answers:

Masja [62]3 years ago

8 0

Answer:

$x=3$

Step-by-step explanation:

Assuming your original equation is

$2ln(e^{ln(5x)})=2ln(15)\\$

by the logarithm property $a*log(b)=log(b^a)$ this equation becomes

$ln(e^{2ln(5x)})=ln(225)$

This means

$e^{2ln(5x)}=225$

Take the natural logarithm of both sides and get:

$2ln(5x)=ln(225)$

By the same lograthim property $a*log(b)=log(b^a)$ , the left side is modified:

$ln(25x^2)=ln(225)$

which gives

$25x^2=225$

$x=\sqrt{\frac{225}{25} }$

$\boxed{x=3}$

ASHA 777 [7]3 years ago

7 0

Answer:

x=3

Step-by-step explanation:

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$\large\boxed{x=\dfrac{-3+\sqrt{40+10\sqrt{10}}}{2}}$

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$2\log x=3-2\log(x+3)\\\\Domain:\ x>0\ \wedge\ x+3>0\to x>-3\\\\D:x>0\\============================\\2\log x=3-2\log(x+3)\qquad\text{add}\ 2\log(x+3)\ \text{to both sides}\\\\2\log x+2\log(x+3)=3\qquad\text{divide both sides by 2}\\\\\log x+\log(x+3)=\dfrac{3}{2}\qquad\text{use}\ \log_ab+\log_ac=\log_a(bc)\\\\\log\bigg(x(x+3)\bigg)=\dfrac{3}{2}\qquad\text{use the de}\text{finition of a logarithm}\\\\x(x+3)=10^\frac{3}{2}\qquad\text{use the distributive property}$

$x^2+3x=10^{1\frac{1}{2}}\\\\x^2+3x=10^{1+\frac{1}{2}}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\x^2+3x=10\cdot10^\frac{1}{2}\qquad\text{use}\ \sqrt[n]{a}=a^\frac{1}{n}\\\\x^2+3x=10\sqrt{10}\qquad\text{subtract}\ 10\sqrt{10}\ \text{from both sides}\\\\x^2+3x-10\sqrt{10}=0\\\\\text{Use the quadratic formula}\\\\ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\a=1,\ b=3,\ c=-10\sqrt{10}\\\\b^2-4ac=3^2-4(1)(-10\sqrt{10})=9+40\sqrt{10}\\\\x=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2(1)}=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2}\\\\x=\dfrac{-3-\sqrt{10+10\sqrt{10}}}{2}\notin D$