What are the potential solutions of In(x^2-25)=0?

Mathematics

1 answer:

Maslowich2 years ago

6 0

Answer:

$x =\pm \sqrt{26}$

Step-by-step explanation:

Given :-

ln ( x² - 25 ) = 0

And we need to find the potential solutions of it. The given equation is the logarithm of x² - 25 to the base e . e is Euler's Number here. So it can be written as ,

Equation :-

$\implies log_e {(x^2-25)}= 0$

In general :-

If we have a logarithmic equation as ,

$\implies log_a b = c$

Then this can be written as ,

$\implies a^c = b$

In a similar way we can write the given equation as ,

$\implies e^0 = x^2 - 25$

Now also we know that $a^0 = 1$ Therefore , the equation becomes ,

$\implies 1 = x^2 - 25 \\\\\implies x^2 = 25 + 1 \\\\\implies x^2 = 26 \\\\\implies x =\sqrt{26} \\\\\implies x = \pm \sqrt{ 26}$

Hence the Solution of the given equation is ±√26.

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BRAINLIEST AWARD NO.2

zzz [600]

Answer:

A whole number first term to render as fifth term a value larger than 10000, should be at least 121

Step-by-step explanation:

The formula is given as recursive since it involves the previous number of the sequence, and defined as:

$a_n=a_{n-1}*3+6$

we also know that the first term is 4

Then in this case, the first five terms are:

$a_1=4\\a_2=4*3+6=18\\a_3=18*3+6=60\\a_4=60*3+6=186\\a_5=186*3+6=564\\$

So if we want to find the first term in the case that the fifth one is greater than 10,000 using this recursive formula, now we have to start backwards, and say that the fifth term is "> 10000" and what the fourth one is.

Notice that if you have this definition for the nth term, we can obtain from it, what the previous term is to find the general rule:

$a_n=a_{n-1}*3+6\\a_n-6=a_{n-1}*3\\\frac{a_n-6}{3} = a_{n-1}\\a_{n-1}=\frac{a_n}{3} -2$

So the rule is to subtract 6 from he term, and divide the subtraction by 3. Then working backwards:

$a_5>10000\\\frac{a_5}{3} -2>\frac{10000}{3} -2\\a_4>=\frac{10000}{3} -2\\\frac{a_4}{3} -2>\frac{\frac{10000}{3}-2}{3}-2 =\frac{10000}{9}-\frac{8}{3} \\a_3>\frac{10000}{9}-\frac{8}{3} \\\frac{a_3}{3} -2>\frac{\frac{10000}{9}-\frac{8}{3} }{3} -2=\frac{10000}{27} -\frac{8}{9} -2=\frac{10000}{27} -\frac{26}{9}\\a_2=\frac{10000}{27} -\frac{26}{9}\\\frac{a_2}{3} -2>\frac{\frac{10000}{27} -\frac{26}{9}}{3} -2=\frac{10000}{81} -\frac{80}{27} \\a_1>\frac{10000}{81} -\frac{80}{27}\approx 120.49$