Question

If (8)^x +1= 64 what is value of x?​

Answer 1

To find : value of $x$

Given : $8^{x+ 1}= 64$

Solution :

• According to  question we have to find the value of  $x$ in the given  expression .
• In $8^{x+ 1}= 64$  we know that $64$ is the square of $8$ . hence we can replace $64$ by $8 ^{2}$ .
• Therefore we get ,

• $8^{x+ 1}= 64$  

       $8^{x + 1 } =8^{2}$

• Now , bases are same of LHS and RHS therefore we compare powers of both the sides and solve for $x$  .

        $x + 1 = 2 \\x = 2-1 \\x = 1$

Hence , value of $x$ is $1$ .

Note : we took $x + 1$ as power of $8$ to solve for $x$  otherwise it would be incorrect question  .

Answer 2

Step-by-step explanation:

$\tt(8) {}^{x + 1} = 64$

$\tt\implies(8) {}^{x + 1} = 8 {}^{2}$

$\tt\implies \: x + 1 = 2$

$\tt\implies \: x = 2 - 1$

$\boxed{ \red {\tt\implies x = 1}}$