What is the answer to 1+log_{4}(t)=6log_{t}(16) ?

The answer to 1+log_{4}(t)=6log_{t}(16) is t=64,t= 1/256

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1+log_{4}(t)=6log_{t}(16)

1 + lo g_{4} (t) = 6 lo g_{t} (16)

t = 64, t = \frac{1}{256}

Decimal

t = 64, t = 0.00390625

Solution steps

1 + lo g_{4} (t) = 6 lo g_{t} (16)

Apply log rules

1 + lo g_{4} (t) = \frac{12}{lo g _{4} ( t )}

Rewrite the equation with

lo g_{4} (t) = u

1 + u = \frac{12}{u}

Solve

1 + u = \frac{12}{u} : u = 3, u = - 4

u = 3, u = - 4

Substitute back

u = lo g_{4} (t),

solve for

t

Solve

lo g_{4} (t) = 3 : t = 64

Solve

lo g_{4} (t) = - 4 : t = \frac{1}{256}

t = 64, t = \frac{1}{256}

Verify Solutions:

t = 64

True

, t = \frac{1}{256}

True

The solutions are

t = 64, t = \frac{1}{256}

x + ln (x) = 10

lo g_{5} (125 x) = \frac{5}{3}

ln (3 x - 4) - ln (x - 3) = ln (2 x)

ln (\frac{t - 63}{125}) = - 0.07 t

9^{l o g_{3} (x)} x^{3} = 243

What is the answer to 1+log_{4}(t)=6log_{t}(16) ?
The answer to 1+log_{4}(t)=6log_{t}(16) is t=64,t= 1/256
Solve for t: 1+log_{4}(t)=6log_{t}(16)
The solution is t=64,t= 1/256