Answer :

We start with the equation
[tex]$$
27^{x-6} = 9^{x-7}.
$$[/tex]

Both 27 and 9 can be written as powers of 3. In fact, we have
[tex]$$
27 = 3^3 \quad \text{and} \quad 9 = 3^2.
$$[/tex]

Substituting these into the equation yields
[tex]$$
(3^3)^{x-6} = (3^2)^{x-7}.
$$[/tex]

Using the law of exponents [tex]$\left(a^b\right)^c = a^{bc}$[/tex], we simplify both sides:
[tex]$$
3^{3(x-6)} = 3^{2(x-7)}.
$$[/tex]

Since the bases are the same, the exponents must be equal for the equality to hold:
[tex]$$
3(x-6) = 2(x-7).
$$[/tex]

Expanding both sides gives:
[tex]$$
3x - 18 = 2x - 14.
$$[/tex]

Subtract [tex]$2x$[/tex] from both sides to collect the [tex]$x$[/tex] terms:
[tex]$$
x - 18 = -14.
$$[/tex]

Then add 18 to both sides:
[tex]$$
x = 4.
$$[/tex]

Thus, the solution to the equation is
[tex]$$
\boxed{4}.
$$[/tex]

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Rewritten by : Barada