Expanding exponents as opposed to solving logarithmically provides different answers. $10^{2t-3} = 7$ The solution should be simplified from the above
$$10^{2t-3} = 7$$
by recognizing the equivalent logarithmic format
$$\log(7) = 2t-3$$
which solves for $t$ as
$$\frac{\log(7)+3}{2} = t$$
Where I'm having problems is I attempted to work the problem from the other direction (just what stood out to me at first) by expanding
$$10^{2t-3} = 10^2 \cdot 10^t \cdot 10^{-3}$$
I think that might be where my problem is, because my next steps solve as
$$100 \cdot 10^t \cdot \frac{1}{1000} = 7$$
$$\frac{100 \cdot 10^t}{1000} = 7$$
$$\frac{1 \cdot 10^t}{10} = 7$$
$$10^t=70$$
Then I apply the equivalent logarithmic format as
$$\log(70) = t$$
and after calculating, it is very clear to me that
$$\log(70) \not = t = \frac{\log(7)+3}{2}$$
Can anyone help me understand what I'm doing wrong here?

$10^{2t}$ does not equal $10^{2} * 10^{t}$.  That would be $10^{t+2}$ instead.
