Question

solve the logarithmic equation using properties of logs. Round to the nearest thousandth if necessary log 2x+logx=2

Answer 1

Given equation is
`\log(2x)+\log(x)=2`

Now we have to solve this equation using properties of logs.

`\log(2x)+\log(x)=2`

Apply formula

`\log(A)+\log(B)=\log(A*B)`

`\log(2x*x)=2`

`\log(2x^2)=2`

convert into exponent form using formula

`\log(a)=b => 10^b=a`

`2x^2=10^2`

`2x^2=100`

`x^2=(100)/(2)`

`x^2=50`

take square root of both sides

x=7.0710678118

Rounding to nearest thousandth gives final answer as x=7.071

Answer 2

It is given that

`log2x + logx =2`

Using
`loga+logb = log(ab)`

We get,
`log(2x^2) = 2`

Also, if
`loga = b` . Then
`a=antilog(b)`

Therefore,
`2x^2 = antilog(2)`

`2x^2 = 100`

Dibide both sides by 2

`x^2 = 50`

`x= \sqrt50`

`x = 7.0711` is the required answer.