Question

1/2 log(5x-4)+ log√x+1=2+log0.18

Answer 1

Answer:

x = 8

Step-by-step explanation:

I guess this equation needs to be solved

Given that

$\frac{1}{2}\log(5x-4)+\log(\sqrt{x+1}) =2+\log0.18$

$\implies \frac{1}{2}\log(5x-4)+\frac{1}{2} \log({x+1) =\log100+\log0.18$

$\implies \frac{1}{2}[\log(5x-4)+ \log({x+1)] =\log(100\times0.18)$

$\implies \frac{1}{2}[\log(5x-4)({x+1)] =\log(18)$

$\implies \log(5x^2-4x+5x-4) =2\log(18)$

$\implies \log(5x^2-4x+5x-4) =\log18^2$

Taking antilog on both sides

$\implies5x^2+x-4 =324$

$\implies5x^2+x-328 =0$

$\implies5x^2+41x-40x-328 =0$

$\implies x(5x+41)-8(5x+41) =0$

$\implies (x-8)(5x+41) =0$

$\implies x=8, -41$

But since the quantity inside the square root should be defined

∴ ${x+1} >0$

$\implies x>-1$

Hence we have only one solution i.e. x = 8

Answer 2

Answer:

x hoga 3

Step-by-step explanation:

x =

Step-by-step explanation:

I guess this equation needs to be solved

Given that

\frac{1}{2}\log(5x-4)+\log(\sqrt{x+1}) =2+\log0.18

2

1

log(5x−4)+log(

x+1

)=2+log0.18

\implies \frac{1}{2}\log(5x-4)+\frac{1}{2} \log({x+1) =\log100+\log0.18

\implies \frac{1}{2}[\log(5x-4)+ \log({x+1)] =\log(100\times0.18)

\implies \frac{1}{2}[\log(5x-4)({x+1)] =\log(18)

\implies \log(5x^2-4x+5x-4) =2\log(18)⟹log(5x

2

−4x+5x−4)=2log(18)

\implies \log(5x^2-4x+5x-4) =\log18^2⟹log(5x

2

−4x+5x−4)=log18

2

Taking antilog on both sides

\implies5x^2+x-4 =324⟹5x

2

+x−4=324

\implies5x^2+x-328 =0⟹5x

2

+x−328=0

on

\implies5x^2+41x-40x-328 =0⟹5x

2

+41x−40x−328=0

\implies x(5x+41)-8(5x+41) =0⟹x(5x+41)−8(5x+41)=0

\implies (x-8)(5x+41) =0⟹(x−8)(5x+41)=0

\implies x=8, -41⟹x=8,−41

But since the quantity inside the square root should be defined

∴ {x+1} > 0x+1>0

\implies x > -1⟹x>−1

Hence we have only one solution i.e. x =