If the given equation is
[tex]27^{\frac13} = 9^{14} + 3^{x + 1}[/tex]
first simplify the left side using the fact that 27 = 3³:
[tex]27^{\frac13} = (3^3)^{\frac13} = 3^{\frac33} = 3^1 = 3[/tex]
and on the right side, 9 = 3²:
[tex]9^{14} = (3^2)^{14} = 3^{2\times14} = 3^{28}[/tex]
So we have
[tex]3 = 3^{28} + 3^{x + 1}[/tex]
Next,
[tex]3^{x+1} = 3^x \times 3^1 = 3 \times 3^x[/tex]
so that
[tex]3 = 3^{28} + 3 \cdot 3^x[/tex]
and dividing both side by 3 gives
[tex]1 = 3^{27} + 3^x[/tex]
Isolate 3ˣ :
[tex]3^x = 1 - 3^{27}[/tex]
Solve for x by taking the base-3 logarithm of both sides:
[tex]\log_3(3^x) = \log_3(1-3^{27})[/tex]
[tex]x \log_3(3) = \log_3(1-3^{27})[/tex]
[tex]\boxed{x = \log_3(1-3^{27})}[/tex]
