Question

Please help me! Its for my big test tomorrow!

Answer 1

QUESTION 11

Given :
`\ln(3x-8)=\ln(x+6)`

We take antilogarithm of both sides to get:

`3x-8=x+6`

Group similar terms to get:

`3x-x=6+8`

Simplify both sides to get:

`2x=14`

Divide both sides by 2 to obtain:

`x=7`

12. Given;
`\log_3(9x-2)=\log_3(4x+3)`

We take antilogarithm to obtain:

`(9x-2)=(4x+3)`

Group similar terms to get:

`9x-4x=3+2`

`5x=5`

We divide both sides by 5 to get:

`x=1`

13.
`\log(4x+1)=\log25`

We take antilogarithm to get:

`(4x+1)=25`

Group similar terms

`4x=25-1`

`4x=24`

Divide both sides by 4

`x=6`

14. Given ;
`\log_6(5x+4)=2`

We take antilogarithm to get:

`(5x+4)=6^2`

Simplify:

`(5x+4)=36`

`5x=36-4`

`5x=32`

Divide both sides by 5

`x=(32)/(5)`

Or

`x=6(2)/(5)`

15. Given:
`\log(10x-7)=3`

We rewrite in the exponential form to get:

`(10x-7)=10^3`

`(10x-7)=1000`

`10x=1000+7`

`10x=1007`

Divide both sides by 10

`x=(1007)/(10)`

16. Given:
`\log_3(4x+2)=\log_3(6x)`

We take antilogarithm to obtain:

`(4x+2)=(6x)`

`2=6x-4x`

Simplify

`2=2x`

Divide both sides by 2

`1=x`

17. Given
`\log_2(3x+12)=4`.

We rewrite in exponential form:

`(3x+12)=2^4`

`(3x+12)=16`

`3x=16-12`

`3x=4`

Divide both sides by 3

`x=(4)/(3)`

18. Given
`\log_3(3x+7)=\log_3(10x)`

We take antilogarithm to get:

`(3x+7)=(10x)`

Group similar terms:

`7=10x-3x`

`7=7x`

We divide both sides by 7

`x=1`

19. Given:
`\log_2x+\log_2(x-3)=2`

Apply the product rule to simplify the left hand side

`\log_2x(x-3)=2`

We take antilogarithm to obtain:

`x(x-3)=2^2`

`x^2-3x=4`

`x^2-3x-4=0`

`(x-4)(x+1)=0`

x=-1 or x=4

But x>0, therefore x=4

20. Given
`\ln x+ \ln (x+4)=3`

Apply product rule to the LHS

`\ln x(x+4)=3`

Rewrite in the exponential form to get:

`x(x+4)=e^3`

`x^2+4x=e^3`

`x^2+4x-e^3=0`

This implies that:

`x=-6.91` or
`x=2.91`