Question

The sum of all the integral roots of (log5 x)2 + log5x(5/×)
= 1 ​

Answer 1

x = 1 ;  x = 5 ;  x = 1 / 25 ;

Given:

• $(log5 x)^{2}$ + $log_{5x}$   (5 / x ) = 1 ​

To Find:

• The sum of all the integral roots of the given equation.

Solution:

           $(log5 x)^{2}$ + $log_{5x}$   (5 / x ) = 1 ​

   

            $(log5 x)^{2}$ +  1 / log5 5x - 1 / logx 5x

           $(log5 x)^{2}$ + 1 / log5 5 + log 5 x  -  1 / logx 5 + logx x   = 1 ​

          let   log5 x = t,

        t2 + 1 / 1+t - 1 / (1/t) + 1    = 1

        t ( t2 + t -2 ) = 0

 =) t = 0                                 ∴ t2 + t - 2 = 0

                                        =)     t2 + 2t - t -2 = 0

                                        =)    ( t - 1 ) ( t + 2 ) = 0

log5 x = 0                       log5 x = 1                           log5 x = -2

x = 1                                   x = 1                                 x = (5)^ -2

                                                                                 x =  1 / 25

Therefore,

x = 1 ;

x = 5 ;

x = 1 / 25 ;

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Answer 2

Answer:

The required answer is 151/25.

Step-by-step explanation:

Rules of the logarithm are:-

• $log a + log b = log(ab)$
• $log a - log b = log(\frac{a}{b})$
• $log_a(b)=\frac{1}{log_b(a)}$

Given:-

$(log_5(x))^2 + log_{5x}(\frac{5}{x} )=1$

To find:-

The sum of all the integral roots of the expression $(log_5(x))^2 + log_{5x}(\frac{5}{x} )=1$.

Step 1 of 3

Consider the given logarithm function as follows:

$(log_5(x))^2 + log_{5x}(\frac{5}{x} )=1$ _____ (1)

Using the rules of logarithm, simplify the expression (1) as follows:

⇒ $(log_5(x))^2 + log_{5x}(5 )-log_{5x}(x)=1$

⇒ $(log_5(x))^2 +\frac{1}{ log_{5}(5x )}-\frac{1}{log_{x}(5x)}=1$ (Since $log_a(b)=\frac{1}{log_b(a)}$)

⇒ $(log_5(x))^2 +\frac{1}{ log_{5}(5 ) + log_{5}(x)}-\frac{1}{log_{x}(5)+log_{x}(x)}=1$

⇒ $(log_5(x))^2 +\frac{1}{ 1 + log_{5}(x)}-\frac{1}{1/(log_{5}(x))+1}=1$ (Since $log_x(x)=1$)

⇒ $(log_5(x))^2 +\frac{1}{ 1 + log_{5}(x)}-\frac{log_{5}(x)}{1+log_{5}(x)}=1$ ____ (2)

Now,

Let $log_{5}(x)=t$.

Then the expression (2) becomes,

$t^2+\frac{1}{1+t}-\frac{t}{1+t}=1$

Further, simplify as follows:

$t^2+\frac{1-t}{1+t}=1$

$t^2(1+t)+1-t=1(1+t)$

$t^2+t^3+1-t=1+t$

$t^3+t^2-2t=0$ ____ (3)

Step 2 of 3

Find the roots of the cubic equation (3) as follows:

$t(t^2+t-2)=0$

Using middle-term splitting method, we have

$t(t^2+2t-t-2)=0$

$t(t(t+2)-1(t+2))=0$

$t(t+2)(t-1)=0$

$t=0$, $t=1$ and $t=-2$

1) When t = 0,

Then, $log_{5}(x)=0$ (Since $log_{5}(x)=t$)

⇒ $5^0=x$

⇒ $x=1$

2) When t = 1,

Then, $log_{5}(x)=1$ (Since $log_{5}(x)=t$)

⇒ $5^1=x$

⇒ $x=5$

3) When t = -2,

Then, $log_{5}(x)=-2$ (Since $log_{5}(x)=t$)

⇒ $5^{-2}=x$

⇒ $x=\frac{1}{5^2}$

⇒ $x=1/25$

Step 3 of 3

Find the $sum$ of all the integral roots.

Since the roots are 1, 5 and 1/25.

Then, the sum is,

= 1 + 5 + 1/25

= $\frac{25+125+1}{25}$

= $\frac{151}{25}$

Final answer: The required answer is 151/25.

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