What is solvefor t,(3/2)^{t-x^2}>(2/3)^{3x-1} ?

The solution to solvefor t,(3/2)^{t-x^2}>(2/3)^{3x-1} is t>-3x+1+x^2

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solvefor t,(3/2)^{t-x^2}>(2/3)^{3x-1}

solve for

t, (\frac{3}{2})^{t - x^{2}} > (\frac{2}{3})^{3 x - 1}

t > - 3 x + 1 + x^{2}

Solution steps

(\frac{3}{2})^{t - x^{2}} > (\frac{2}{3})^{3 x - 1}

Apply exponent rules

t - x^{2} > - 1 \cdot (3 x - 1)

Simplify

t - x^{2} > - (3 x - 1)

Move

x^{2}

to the right side

t > - 3 x + 1 + x^{2}

(\frac{1}{5})^{2 x^{2} - 3 x} \geq 5

(2^{x})^{2} \leq 1 5^{x}

\frac{1}{2 ^{x} - 1} > \frac{1}{1 - 2 ^{x - 1}}

((\frac{3}{7})^{x})^{2} < (\frac{7}{3})^{4} x - 2

(\frac{2}{5})^{\frac{( 1 - 5 x )}{( 4 + 16 )}} < \frac{25}{4}

What is solvefor t,(3/2)^{t-x^2}>(2/3)^{3x-1} ?
The solution to solvefor t,(3/2)^{t-x^2}>(2/3)^{3x-1} is t>-3x+1+x^2