Question

Logbase11(2x-1) = 1 - logbase11 (x+4). find the value of x with detailed reasoning?

Answer 1

`x=3/2`

Explanation:

Note that in `log_11(2x-1)=1-log_11(x+4)`, we cannot have `2x-1<0` and `x+4<0` i.e. the domain is `x>=1/2`.

We can write `log_11(2x-1)=1-log_11(x+4)` as

`log_11(2x-1)+log_11(x+4)=1`

i.e. `log_11[(2x-1)(x+4)]=1`

or `(2x-1)(x+4)=11`

or `2x^2+8x-x-4=11`

or `2x^2+7x-15=0`

or `2x^2-3x+10x-15=0`

or `x(2x-3)+5(2x-3)=0`

or `(x+5)(2x-3)=0`

i.e. either `x+5=0` or `2x-3=0`

i.e. `x=-5` or `x=3/2`

but as `x=-5` is not in the domain.

hence `x=3/2`

Answer 2

`x=3/2`

Explanation:

`log_(11)(2x-1)=1-log_(11)(x+4)`

Add `bb(log_(11)(x+4))` to both sides:

`log_(11)(2x-1)+log_(11)(x+4)=1`

By the law of logarithms:

`loga+logb=log(ab)`

`:.`

`log_(11)(2x-1)+log_(11)(x+4)=log_(11)((2x-1)(x+4))`

`-> log_(11)(2x^2+7x-4)=1`

By the law of logarithms:

`b^(log_ba)=a`

`:.`

`11^(log_(11)(2x^2+7x-4))=11^1`

`-> 2x^2+7x-4=11`

`2x^2+7x-15=0`

Solving for `x`

Factor:

`(x+5)(2x-3)=0=>x=-5 and x= 3/2`

For real numbers:

`log(x)` only exist if `x>0`

For:

`x=-5`

`(2x-1)=(2(-5)-1)=-11color(white)(88)` Invalid for real numbers.

For:

`x=3/2`

`(2x-1)=(2(3/2)-1)=2`

and:

`(x+4)=((3/2)+4)=11/2`

Only `x=3/2`