2.8=-log_{10}(h^{30})

Solución

2.8 = - lo g_{10} (h^{30})

h = \frac{1.62657 E 15}{2.5 E 15}, h = - \frac{1.62657 E 15}{2.5 E 15}

Notación decimal

h = 0.80661 \dots, h = - 0.80661 \dots

Pasos de solución

2.8 = - lo g_{10} (h^{30})

Mover

lo g_{10} (h^{30})

al lado izquierdo

2.8 + lo g_{10} (h^{30}) = 0

Mover

2.8

al lado derecho

lo g_{10} (h^{30}) = - 2.8

Aplicar las leyes de los exponentes:

a^{2 \cdot n} = (a^{2})^{n}

h^{30} = (h^{2})^{15}

lo g_{10} ((h^{2})^{15}) = - 2.8

Aplicar las propiedades de los logaritmos

h^{2} = \frac{1}{1 0 ^{0.18666 \dots}}

Resolver

h^{2} = \frac{1}{1 0 ^{0.18666 \dots}} : h = \frac{1.62657 E 15}{2.5 E 15}, h = - \frac{1.62657 E 15}{2.5 E 15}

h = \frac{1.62657 E 15}{2.5 E 15}, h = - \frac{1.62657 E 15}{2.5 E 15}

Verificar las soluciones:

h = \frac{1.62657 E 15}{2.5 E 15}

Verdadero

, h = - \frac{1.62657 E 15}{2.5 E 15}

Verdadero

Las soluciones son

h = \frac{1.62657 E 15}{2.5 E 15}, h = - \frac{1.62657 E 15}{2.5 E 15}

2 ln (x) - 2 ln (x + 2) = 4 ln (x^{2})

ln (x^{2} - x - 1) = 0

lo g_{3} (2 x^{2} + 5 x + 4) - 7 = 2

\frac{lo g _{10} ( 6 - x ^{2} )}{lo g _{10} ( 3 x - 4 )} = 2

lo g_{4} (x + 12) - lo g_{4} (5 - x) = 2