Please solve both of these questions ASAP urgent need

1) Solve : log₃ ( 1 + log₃ ( 2^x - 7 ) ) = 1

2) Solve: log_1/4 ( 2 - x ) > log_{1/4 $(\frac{2}{x + 1})$}

Question

Please solve both of these questions ASAP urgent need

1) Solve : log3 ( 1 + log3 ( 2x -
7 ) ) = 1

2) Solve: log1/4 ( 2 - x ) > log1/4 2 x   +
  1

Lovina Kansal · Accepted Answer

Dear student
Que 1
We have,log31+log32x-7=1Take 2x=uSo, log31+ log3u-7=1⇒ log31+ log3u-7= log33⇒1+ log3u-7=3⇒ log3u-7=2⇒ log3u-7= log332    As a= logbba⇒ log3u-7= log39⇒u-7=9⇒u=16⇒2x=16⇒2x=24On comparing, we get x=4
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Please solve both of these questions ASAP urgent need 1) Solve : log3 ( 1 + log3 ( 2x - 7 ) ) = 1 2) Solve: log1/4 ( 2 - x ) > log1/4 2 x + 1

Please solve both of these questions ASAP urgent need

1) Solve : log₃ ( 1 + log₃ ( 2^x - 7 ) ) = 1

2) Solve: log_1/4 ( 2 - x ) > log_{1/4 $(\frac{2}{x + 1})$}