Question

please answer fast, it's urgent ​

Answer 1

Required Answer:-

Given:

• $\rm {x}^{1 - \log_{5}(x) } = 0.04$

To find:

• The value of x.

Solution:

Given that,

$\rm \implies {x}^{1 - \log_{5}(x) } = 0.04$

$\rm \implies {x}^{1 - \log_{5}(x) } = \dfrac{4}{100}$

$\rm \implies {x}^{1 - \log_{5}(x) } = \dfrac{1}{25}$

Taking logarithm of base 5 on both sides, we get,

$\rm \implies \log_{5} \big({x}^{1 - \log_{5}(x) } \big) = log_{5} \bigg( \dfrac{1}{25} \bigg)$

$\rm \implies (1 - log_{5}(x) ) \log_{5}(x) = log_{5}( {5}^{ - 2} )$

$\rm \implies (1 - log_{5}(x) ) \log_{5}(x) = - 2log_{5}(5)$

$\rm \implies (1 - log_{5}(x) ) \log_{5}(x) = - 2$

Let us assume that,

$\rm \implies y = \log_{5}(x)$

Therefore,

$\rm \implies (1 - y)y = - 2$

$\rm \implies - {y}^{2} + y = - 2$

$\rm \implies {y}^{2} - y - 2 = 0$

$\rm \implies {y}^{2} - 2y + y- 2 = 0$

$\rm \implies y(y- 2) + 1(y- 2 )= 0$

$\rm \implies (y+ 1)(y- 2 )= 0$

By zero product rule,

Either y + 1 = 0 or y - 2 = 0

Therefore,

➡ y = -1, 2

So, when y = -1

$\rm \implies \log_{5}(x) = - 1$

$\rm \implies x = {5}^{ - 1}$

$\rm \implies x = \dfrac{1}{5}$

Again, when y = 2

$\rm \implies \log_{5}(x) = 2$

$\rm \implies x = {5}^{2}$

$\rm \implies x =25$

Hence, the possible values of x are 1/5 and 25.

Answer:

• The possible values of x are 1/5 and 25.

Answer 2

$\rm{x^{1-\log _5\left(x\right)}=0.04}$

• Apply Exponent Rules

$\to\rm{\displaystyle e^{\left(1-\log _5\left(x\right)\right)\ln \left(x\right)}=0.04}$

$\to\rm{\displaystyle \ln \left(e^{\left(1-\log _5\left(x\right)\right)\ln \left(x\right)}\right)=\ln \left(0.04\right)}$

$\to\rm{\displaystyle \left(1-\log _5\left(x\right)\right)\ln \left(x\right)\ln \left(e\right)=\ln \left(0.04\right)}$

$\to\rm{\displaystyle \left(1-\log _5\left(x\right)\right)\ln \left(x\right)=\ln \left(0.04\right)}$

• Apply Log Rules

$\to\rm{\displaystyle \left(1-\log _5\left(x\right)\right)\frac{\log _5\left(x\right)}{\log _5\left(e\right)}=\ln \left(0.04\right)}$

$\to\rm{\displaystyle \frac{\log _5\left(x\right)\left(1-\log _5\left(x\right)\right)}{\log _5\left(e\right)}=\ln \left(0.04\right)}$

• Rewrite The Equation With log₅(x) = u

$\to\rm{\displaystyle \frac{u\left(1-u\right)}{\log _5\left(e\right)}=\ln \left(0.04\right)}$

$\to\rm{\displaystyle \frac{u\left(1-u\right)}{\log _5\left(e\right)}\log _5\left(e\right)=\ln \left(0.04\right)\log _5\left(e\right)}$

$\to\rm{\displaystyle u\left(1-u\right)=\frac{\ln \left(0.04\right)}{\ln \left(5\right)}}$

$\to\rm{\displaystyle u-u^2=\frac{\ln \left(0.04\right)}{\ln \left(5\right)}}$

$\to\rm{\displaystyle u-u^2-\frac{\ln \left(0.04\right)}{\ln \left(5\right)}=\frac{\ln \left(0.04\right)}{\ln \left(5\right)}-\frac{\ln \left(0.04\right)}{\ln \left(5\right)}}$

$\to\rm{\displaystyle -u^2+u+2=0}$

• Solve with the quadratic formula

$\to\rm{\displaystyle u_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\left(-1\right)\cdot \:2}}{2\left(-1\right)}}$

$\to\rm{\displaystyle u_{1,\:2}=\frac{-1\pm \:3}{2\left(-1\right)}}$

• Separate the Solutions

$\to\rm{\displaystyle u_{1}=\frac{-1+3}{2(-1)}, u_{2}=\frac{-1-3}{2(-1)}}$

• The solutions to he quadratic equation are

$\to\rm{\displaystyle u=-1,\:u=2}$

• Substitute Back u = log₅(x) , Solve For x

$\boxed{\to\rm{x=\frac{1}{5},\:x=25}}$