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

$x=\frac{32}{5}$

Or

$x=6\frac{2}{5}$

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

$x=\frac{1007}{10}$

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

$x=\frac{4}{3}$

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 $x=2.91$