What is the answer to solvefor a,1=blog_{5}(6+a) ?

The answer to solvefor a,1=blog_{5}(6+a) is a=5^{1/b}-6 {b 0}

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solvefor a,1=blog_{5}(6+a)

solve for

a, 1 = b lo g_{5} (6 + a)

a = 5^{\frac{1}{b}} - 6 {b < 0 or b > 0}

Solution steps

1 = b lo g_{5} (6 + a)

Switch sides

b lo g_{5} (6 + a) = 1

Divide both sides by

b

lo g_{5} (6 + a) = \frac{1}{b}

Apply log rules

6 + a = 5^{\frac{1}{b}}

Solve

6 + a = 5^{\frac{1}{b}} : a = 5^{\frac{1}{b}} - 6

a = 5^{\frac{1}{b}} - 6

Verify Solutions:

a = 5^{\frac{1}{b}} - 6 {b < 0 or b > 0}

The solution is

a = 5^{\frac{1}{b}} - 6 {b < 0 or b > 0}

ln (3 x + 4) = 6

ln (2 x + 1) + ln (x + 2) = 0

lo g_{2} (\frac{1}{x ^{2}}) = 3

5.5 = 6 + lo g_{10} (x)

lo g_{2} (x - 3) + lo g_{2} (2 x - 1) = 0

What is the answer to solvefor a,1=blog_{5}(6+a) ?
The answer to solvefor a,1=blog_{5}(6+a) is a=5^{1/b}-6 {b<0\lor b>0}