What is the answer to 2ln(X)-3=ln(A^2-e^3) ?

The answer to 2ln(X)-3=ln(A^2-e^3) is X=e^{(ln(A^2-e^3)+3)/2} {A esqrt(e)}

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2ln(X)-3=ln(A^2-e^3)

2 ln (X) - 3 = ln (A^{2} - e^{3})

X = e^{\frac{l n ( A ^{2} - e ^{3} ) + 3}{2}} {A < - e e or A > e e}

Solution steps

2 ln (X) - 3 = ln (A^{2} - e^{3})

Move

3

to the right side

2 ln (X) = ln (A^{2} - e^{3}) + 3

Divide both sides by

2

ln (X) = \frac{ln ( A ^{2} - e ^{3} ) + 3}{2}

Apply log rules

X = e^{\frac{l n ( A ^{2} - e ^{3} ) + 3}{2}}

Verify Solutions:

X = e^{\frac{l n ( A ^{2} - e ^{3} ) + 3}{2}} {A < - e e or A > e e}

The solution is

X = e^{\frac{l n ( A ^{2} - e ^{3} ) + 3}{2}} {A < - e e or A > e e}

20 \cdot lo g_{10} (x \cdot 2) = 20

x, 4 ln (8 x - 2) + 2 = 10

lo g_{8} (a^{6}) = lo g_{8} (x)

8 + lo g_{6} (- 4 x) = 10

ln (x - 4) = 2 - ln (x + 4)

What is the answer to 2ln(X)-3=ln(A^2-e^3) ?
The answer to 2ln(X)-3=ln(A^2-e^3) is X=e^{(ln(A^2-e^3)+3)/2} {A<-esqrt(e)\lor A>esqrt(e)}