Question

Write your question here (Keep it simple and clear to get the best answer) if log (2x+1)-log (3x -2)=1

Answer 1

Question:

Find the value of "x" if ;

log(2x+1) - log (3x-2) = 1.

Answer:

x = 3/4

Note:

• logA + logB = log(A•B)

• logA - logB = log(A/B)

• B•logA = log(A^B)

Solution:

We have;

=> log(2x+1) - log (3x-2) = 1

=> log{(2x+1)/(3x-2)} = 1

{ considering the base 10 }

=> (2x+1)/(3x-2) = 10^1

=> (2x+1)/(3x-2) = 10

=> 2x + 1 = 10•(3x-2)

=> 2x + 1 = 30x - 20

=> 30x - 2x = 20 + 1

=> 28x = 21

=> x = 21/28

=> x = 3/4

Hence,

The required value of "x" is 3/4 .

Answer 2

AnswEr :

$\longrightarrow\sf{ log(2x + 1) - log(3x - 2) = 1}$

• logA + logB = $\sf{ log( \frac{a}{b} ) }$

$\longrightarrow\sf{ log \bigg( \dfrac{2x + 1}{3x - 2} \bigg) = 1}$

• considering the Base as 10

$\longrightarrow\sf{ \dfrac{2x + 1}{3x - 2} = {10}^{1} }$

$\longrightarrow\sf{ \dfrac{2x + 1}{3x - 2} = {10}}$

$\longrightarrow\sf{2x + 1 = 10(3x - 2)}$

$\longrightarrow\sf{2x + 1 = 30x - 20}$

$\longrightarrow\sf{20 + 1 = 30x - 2x}$

$\longrightarrow\sf{21 = 28x}$

$\longrightarrow\sf{x = \dfrac{21}{28} }$

$\longrightarrow\boxed{\sf{x = \dfrac{3}{4} }}$

⠀

$\therefore$ Required Value of x is $\sf{\dfrac{3}{4}}$