What is the answer to ln(γ)=ln(4+e^{2^β})+ω ?

The answer to ln(γ)=ln(4+e^{2^β})+ω is γ=4e^ω+e^{ω+2^β}

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ln(γ)=ln(4+e^{2^β})+ω

ln (γ) = ln (4 + e^{2^{β}}) + ω

γ = 4 e^{ω} + e^{ω + 2^{β}}

Solution steps

ln (γ) = ln (4 + e^{2^{β}}) + ω

Apply log rules

γ = 4 e^{ω} + e^{ω + 2^{β}}

lo g_{7} (\frac{1}{x}) = - 3

lo g_{2} (x + 3) + lo g_{2} (x - 5) = 0

lo g_{5} (x + 6) = - lo g_{5} (x + 2) + 1

lo g_{4} (x + 11) = 2

20 \cdot lo g_{10} (x) = 0

What is the answer to ln(γ)=ln(4+e^{2^β})+ω ?
The answer to ln(γ)=ln(4+e^{2^β})+ω is γ=4e^ω+e^{ω+2^β}