Question

Solve log6 + log6 (x-1) = 1

Answer 1

`\boxed{\sf x = 7 }`

Explanation:

We are here given a logarithmic equation and we need to solve it out and then find the value of x. The given equation is ,

`\sf\longrightarrow log_6 + log_6 ( x -1) = 1`

Here I am assuming that the base of the logarithm is 6 . The equation can be written as ,

`\sf\longrightarrow log_6 1+ log_6 ( x -1) = 1`

Recall the property of log as ,
`\sf log_x a + log_x b = log_x(ab)` , on using this property we have ,

`\sf\longrightarrow log_6 \{ 1 ( x -1)\} = 1`

Simplify ,

`\sf\longrightarrow log_6 ( x -1) = 1`

We know that , if
`\sf log_a b = c` then in expotential form it can be expressed as
`\sf a^c = b` . Using this we have ,

`\sf\longrightarrow 6^1 = x - 1`

Simplify ,

`\sf\longrightarrow 6 = x - 1`

Add 1 both sides ,

`\sf\longrightarrow x = 6 + 1`

Therefore ,

`\sf\longrightarrow \boxed{\blue{\sf \quad x = 7\quad }}`

Hence the value of x is 7 .

Answer 2

x=7

Explanation:

log6(1) + log6 (x-1) = 1

We know that log (a) * log (b) = log (ab)

log6 (1*(x-1)) = 1

log6 (x-1) = 1

Raising each side to the base of 6

6 ^log6 (x-1) = 6^1

x-1 = 6

Add 1 to each sdie

x-1+1 = 6+1

x=7