What is the answer to solvefor t,4((x^2)^{-t})/5 =288 ?

The answer to solvefor t,4((x^2)^{-t})/5 =288 is t=-(ln(360))/(2ln(x))

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solvefor t,4((x^2)^{-t})/5 =288

solve for

t, 4 \frac{( x ^{2} ) ^{- t}}{5} = 288

t = - \frac{ln ( 360 )}{2 ln ( x )}

Solution steps

4 \cdot \frac{( x ^{2} ) ^{- t}}{5} = 288

Apply exponent rules

2 (- t) ln (x) = ln (360)

Solve

2 (- t) ln (x) = ln (360) : t = - \frac{ln ( 360 )}{2 ln ( x )}

t = - \frac{ln ( 360 )}{2 ln ( x )}

2^{x} + 3 \cdot 2^{x} = 16

x, \frac{( 9 ^{x - 1} )}{( 3 ^{y + 2} )} = 81

1 6^{x} + 2 0^{x} + 2 5^{x} = 0

3 2^{x} = 8

(\frac{1}{9})^{6 x + 1} = 9^{7 - x}

What is the answer to solvefor t,4((x^2)^{-t})/5 =288 ?
The answer to solvefor t,4((x^2)^{-t})/5 =288 is t=-(ln(360))/(2ln(x))