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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?

Wayne Roberts
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  • What makes you think $10^{2t} = 10^210^t$ and $10^{2t-3} = 10^{2t}10^{-3}$ can be true simultaneously? – letsintegreat Feb 08 '21 at 04:54
  • @RobertTheTutor & Dhanvi Sreenivasan explained where I had a misunderstanding of exponential rules. I originally was thinking $10^{2t} = 10^2 \cdot 10^t$ as additive when instead I should have simplified it to $(10^2)^t$. – Wayne Roberts Feb 09 '21 at 05:34

2 Answers2

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

RobertTheTutor
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  • A good way to remember the rules is to write out explicitly examples with small powers of 2. Like $\frac{2^5}{2^3} = \frac{22222}{222} =2*2 = 2^2$ – RobertTheTutor Feb 09 '21 at 13:05
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$$10^2\cdot 10^t \cdot 10^{-3} = 10^{2+t-3} = 10^{t-1} \neq 10^{2t-3}$$

Dhanvi Sreenivasan
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