Question

What is the solution of log2x + 727 = 3?

Answer 1

The solution of
`log_(2x+7)27\:=\:3` is -2.

Explanation:

Given :
`log_(2x+7)27\:=\:3`

We have to solve for x.

Consider the given expression
`log_(2x+7)27\:=\:3`

Apply rule,
`\log _a\left(b\right)=(\ln \left(b\right))/(\ln \left(a\right))`

`\log _(2x+7)\left(27\right)=(\ln \left(27\right))/(\ln \left(2x+7\right))`

On simplifying, we get,

`(\ln \left(27\right))/(\ln \left(2x+7\right))=3`

Multiply both side by
`\ln \left(2x+7\right)`

We have,

`(\ln \left(27\right))/(\ln \left(2x+7\right))\ln \left(2x+7\right)=3\ln \left(2x+7\right)`

Multiply fractions as
`\:a\cdot (b)/(c)=(a\:\cdot \:b)/(c)`

we get,

`=(\ln \left(27\right)\ln \left(2x+7\right))/(\ln \left(2x+7\right))==\ln \left(27\right)`

Thus, becomes
`\ln \left(27\right)=3\ln \left(2x+7\right)`

Divide both sides by 3, we have,

`(3\ln \left(2x+7\right))/(3)=(\ln \left(27\right))/(3)`

Apply rule
`\log _a\left(x^b\right)=b\cdot \log _a\left(x\right)`

`\ln(27)=\ln \left(3^3\right)=3\ln \left(3\right)`

Thus,
`\ln \left(2x+7\right)=\ln \left(3\right)`

`\mathrm{When\:the\:logs\:have\:the\:same\:base:\:\:}\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\quad \Rightarrow \quad f\left(x\right)=g\left(x\right)`

2x+ 7 = 3

Subtract 7 both sides, we get,

2x = 3 - 7

Simplify , we have,

2x = -4

Divide both side by 2, we have

x = -2

Thus, the solution of
`log_(2x+7)27\:=\:3` is -2.

Answer 2

`log _(2x+7) 27 = 3`
( 2 x + 7 )³ = 27
( 2 x + 7 )³ = 3³
2 x + 7 = 3
2 x = 3 - 7
2 x = - 4
x = ( - 4 ) : 2
x = - 2