What is the answer to 63=95-93log_{10}(t+1) ?

The answer to 63=95-93log_{10}(t+1) is t=10^{32/93}-1

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63=95-93log_{10}(t+1)

63 = 95 - 93 lo g_{10} (t + 1)

t = 1 0^{\frac{32}{93}} - 1

Decimal Notation

t = 1.20844 \dots

Solution steps

63 = 95 - 93 lo g_{10} (t + 1)

Switch sides

95 - 93 lo g_{10} (t + 1) = 63

Move

95

to the right side

- 93 lo g_{10} (t + 1) = - 32

Divide both sides by

- 93

lo g_{10} (t + 1) = \frac{32}{93}

Apply log rules

t + 1 = 1 0^{\frac{32}{93}}

Solve

t + 1 = 1 0^{\frac{32}{93}} : t = 1 0^{\frac{32}{93}} - 1

t = 1 0^{\frac{32}{93}} - 1

Verify Solutions:

t = 1 0^{\frac{32}{93}} - 1

True

The solution is

t = 1 0^{\frac{32}{93}} - 1

lo g_{a} (64) = 6

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

lo g_{2} (4 x - 9) = 1

lo g_{2} (x) - lo g_{4} (2 x - 1) = 1

x, ln (y) + y = x + ln (x) + c

What is the answer to 63=95-93log_{10}(t+1) ?
The answer to 63=95-93log_{10}(t+1) is t=10^{32/93}-1
Solve for t: 63=95-93log_{10}(t+1)
The solution is t=10^{32/93}-1