What is the answer to 2.8=-log_{10}(h^{30}) ?

The answer to 2.8=-log_{10}(h^{30}) is h=sqrt((1.62657E15)/(2.5E15)),h=-sqrt((1.62657E15)/(2.5E15))

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

The solution is h=sqrt((1.62657E15)/(2.5E15)),h=-sqrt((1.62657E15)/(2.5E15))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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}

Decimal Notation

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

Solution steps

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

Move

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

to the left side

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

Move

2.8

to the right side

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

Apply exponent rule:

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

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

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

Apply log rules

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

Solve

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}

Verify Solutions:

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

True

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

True

The solutions are

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

What is the answer to 2.8=-log_{10}(h^{30}) ?
The answer to 2.8=-log_{10}(h^{30}) is h=sqrt((1.62657E15)/(2.5E15)),h=-sqrt((1.62657E15)/(2.5E15))
Solve for h: 2.8=-log_{10}(h^{30})
The solution is h=sqrt((1.62657E15)/(2.5E15)),h=-sqrt((1.62657E15)/(2.5E15))