Question

Solve and show your steps
3^2x+3 = 27^x-5

Answer 1

To solve the equation \( 3^{2x + 3} = 27^{x - 5} \), we first express \( 27 \) as a power of \( 3 \):

\[
27 = 3^3
\]

Thus, we can rewrite the right side of the equation:

\[
27^{x - 5} = (3^3)^{x - 5} = 3^{3(x - 5)} = 3^{3x - 15}
\]

Now, we can substitute this back into the original equation:

\[
3^{2x + 3} = 3^{3x - 15}
\]

Since the bases are the same, we can set the exponents equal to one another:

\[
2x + 3 = 3x - 15
\]

Now, we will solve for \( x \). First, rearrange the equation to isolate \( x \):

\[
3 + 15 = 3x - 2x
\]

This simplifies to:

\[
18 = x
\]

Thus, the solution is:

\[
\boxed{18}
\]

To verify, we can substitute \( x = 18 \) back into the original equation:

Left side:

\[
3^{2(18) + 3} = 3^{36 + 3} = 3^{39}
\]

Right side:

\[
27^{18 - 5} = 27^{13} = (3^3)^{13} = 3^{39}
\]

Since both sides are equal, the solution \( x = 18 \) is correct.