What is the answer to solvefor a,log_{2}(a^3)+log_{2}(b)=8 ?

The answer to solvefor a,log_{2}(a^3)+log_{2}(b)=8 is a=2^{(8-log_{2}(b))/3} {b>0}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor a,log_{2}(a^3)+log_{2}(b)=8

solve for

a, lo g_{2} (a^{3}) + lo g_{2} (b) = 8

a = 2^{\frac{8 - l o g _{2} ( b )}{3}} {b > 0}

Solution steps

lo g_{2} (a^{3}) + lo g_{2} (b) = 8

Move

lo g_{2} (b)

to the right side

lo g_{2} (a^{3}) = 8 - lo g_{2} (b)

Apply log rules

a = 2^{\frac{8 - l o g _{2} ( b )}{3}}

Verify Solutions:

a = 2^{\frac{8 - l o g _{2} ( b )}{3}} {b > 0}

The solution is

a = 2^{\frac{8 - l o g _{2} ( b )}{3}} {b > 0}

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

2 lo g_{10} (x) = 4

lo g_{3} ((x + 2)^{2}) + 1 = 7

lo g_{10} (x^{2} - 3 x + 1) = lo g_{10} (- 3 x + 10)

x, lo g_{6} (x) + lo g_{6} (y) = 1

What is the answer to solvefor a,log_{2}(a^3)+log_{2}(b)=8 ?
The answer to solvefor a,log_{2}(a^3)+log_{2}(b)=8 is a=2^{(8-log_{2}(b))/3} {b>0}