Question

Solve for x:
log base 6 is (x-2) +log base 6 (x+3)=1

Answer 1

i hope this answer would be helpful for you

Answer 2

The required value of x = 3

Given :

$\displaystyle \sf{ log_{6}(x - 2) + log_{6}(x + 3) = 1}$

To find :

The value of x

Solution :

Step 1 of 2 :

Write down the given equation

Here the given equation is

$\displaystyle \sf{ log_{6}(x - 2) + log_{6}(x + 3) = 1}$

Step 2 of 2 :

Find the value of x

$\displaystyle \sf{ log_{6}(x - 2) + log_{6}(x + 3) = 1}$

$\displaystyle \sf{ \implies log_{6} \{(x - 2) (x + 3) \}= 1}$

$\displaystyle \sf{ \implies (x - 2) (x + 3) = {6}^{1} }$

$\displaystyle \sf{ \implies {x}^{2} + 3x - 2x - 6 = 6 }$

$\displaystyle \sf{ \implies {x}^{2} + x - 12 = 0 }$

$\displaystyle \sf{ \implies {x}^{2} + (4 - 3)x - 12 = 0 }$

$\displaystyle \sf{ \implies {x}^{2} + 4x - 3x - 12 = 0 }$

$\displaystyle \sf{ \implies x(x + 4) - 3(x + 4) = 0 }$

$\displaystyle \sf{ \implies (x + 4) (x - 3) = 0 }$

$\displaystyle \sf{ \implies x = - 4 \:, \: 3 }$

For x = - 4 , log₆ (x - 2) & log₆ (x + 3) is undefined

Hence the required value of x = 3

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