Question

Solve log6 + log6 (x-1) = 1

Answer 1

Answer:

x=7

Step-by-step 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

Answer 2

Answer:

$\boxed{\sf x = 7 }$

Step-by-step 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 .