Question

Given that 27^1/3= 9^14 + 3^x+1 find the exact value of x

Answer 1

If the given equation is

$27^{\frac13} = 9^{14} + 3^{x + 1}$

first simplify the left side using the fact that 27 = 3³:

$27^{\frac13} = (3^3)^{\frac13} = 3^{\frac33} = 3^1 = 3$

and on the right side, 9 = 3²:

$9^{14} = (3^2)^{14} = 3^{2\times14} = 3^{28}$

So we have

$3 = 3^{28} + 3^{x + 1}$

Next,

$3^{x+1} = 3^x \times 3^1 = 3 \times 3^x$

so that

$3 = 3^{28} + 3 \cdot 3^x$

and dividing both side by 3 gives

$1 = 3^{27} + 3^x$

Isolate 3ˣ :

$3^x = 1 - 3^{27}$

Solve for x by taking the base-3 logarithm of both sides:

$\log_3(3^x) = \log_3(1-3^{27})$

$x \log_3(3) = \log_3(1-3^{27})$

$\boxed{x = \log_3(1-3^{27})}$