$$\frac{10^{-4/5} \cdot 10^{1/15}}{10^{2/3}}$$
The answer is $10^{-7/5}$, which seems impossible to me. I get:
$10^{-4/5} \cdot 10^{-11/15}$. I see where the numerator $7$ comes from but the denominator is being a pest, and won't let me do anything because I have to make them equal to add them.
$\endgroup$ 24 Answers
$\begingroup$Note that $$\frac{10^{-4/5} \cdot 10^{1/15}}{10^{2/3}}=10^{-4/5} \cdot\frac{ 10^{1/15}}{10^{10/15}}=10^{-4/5} \cdot10^{(1-10)/15}=10^{-4/5}\cdot10^{-3/5}=\boxed{10^{-7/5}}$$ as desired.
$\endgroup$ $\begingroup$After taking a breather and relaxing, I realized my mistake, a/amn is subtracting, not adding, giving me -9/15, reducing that to -3/5 -4-3 = -7/5
$\endgroup$ $\begingroup$$$10^{-12/15}\;.\;10^{1/15}\; . \;10^{-10/15}$$
$$10^{(-12-10+1)/15}$$
$$10^{-21/15}$$
$$10^{-7/5}$$
$\endgroup$ $\begingroup$It looks like you made a sign error while combining exponents.
$$\frac1{15} - \frac23 = \frac1{15} - \frac{10}{15} = -\frac9{15}.$$
You seem to have gotten $-\frac{11}{15}$ when you should have gotten $-\frac9{15},$ perhaps by flipping the sign of $\frac1{15}.$
Of course $-\frac9{15} = -\frac35,$ and the last step is easy.
The fact that $4 - 11 = -7$ is a complete red herring. If the problem were just a little different so the denominators happened to be equal, for example if they were both $5,$ you would have $$ -\frac45 - \frac{11}5 = -\frac{15}5 = -3.$$ So it seems you have a tendency to make sign errors while adding or subtracting. Now that you know this, you may be able to take steps to correct it.
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