We need to solve the exponential equation $6^{1-T} = 5^{3T+1}$ for the variable $T$.

1. Problem Description

We need to solve the exponential equation $6^{1-T} = 5^{3T+1}$ for the variable $T$.

2. Solution Steps

To solve the exponential equation $6^{1-T} = 5^{3T+1}$, we can take the logarithm of both sides. We can use any base for the logarithm, but it's common to use the natural logarithm (base $e$), denoted as $ln$.

Taking the natural logarithm of both sides gives us:

$ln(6^{1-T}) = ln(5^{3T+1})$

Using the logarithm power rule $ln(a^b) = b \cdot ln(a)$, we have:

$(1-T) \cdot ln(6) = (3T+1) \cdot ln(5)$

Now, we distribute the $ln(6)$ and $ln(5)$ on each side:

$ln(6) - T \cdot ln(6) = 3T \cdot ln(5) + ln(5)$

Next, we want to isolate $T$ terms on one side of the equation and constant terms on the other side. So, we add $T \cdot ln(6)$ to both sides and subtract $ln(5)$ from both sides:

$ln(6) - ln(5) = 3T \cdot ln(5) + T \cdot ln(6)$

Now, we factor out $T$ from the right side:

$ln(6) - ln(5) = T(3ln(5) + ln(6))$

Finally, we divide both sides by $(3ln(5) + ln(6))$ to solve for $T$:

$T = \frac{ln(6) - ln(5)}{3ln(5) + ln(6)}$

We can simplify the expression using logarithm properties. Specifically, $a \cdot ln(b) = ln(b^a)$.

$T = \frac{ln(6) - ln(5)}{ln(5^3) + ln(6)}$

$T = \frac{ln(6) - ln(5)}{ln(125) + ln(6)}$

$T = \frac{ln(6/5)}{ln(125 \cdot 6)}$

$T = \frac{ln(6/5)}{ln(750)}$

We can also approximate the value using a calculator:

$ln(6) \approx 1.791759$

$ln(5) \approx 1.609438$

$3ln(5) \approx 4.828314$

$T \approx \frac{1.791759 - 1.609438}{4.828314 + 1.791759} = \frac{0.182321}{6.620073} \approx 0.02754$

3. Final Answer

$T = \frac{ln(6) - ln(5)}{3ln(5) + ln(6)} = \frac{ln(6/5)}{ln(750)} \approx 0.02754$