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bases: write 3/2 as a power with a base of 10
- Thread starter NEHA
- Start date
~
Hello, NEHA!
We have: \(\displaystyle \:10^x\;=\;1.5\)
Take logs of both sides (base 10): \(\displaystyle \:\log\left(10^x\right) \;= \;\log(1.5)\)
Then we have: \(\displaystyle \:x\cdot\log(10) \;= \;\log(1.5)\)
Since \(\displaystyle \,\log(10)\,=\,1\), we have: \(\displaystyle \:x\;=\;\log(1.5) \;= \;0.176091259...\)
Therefore: \(\displaystyle \L\:\frac{3}{2} \:\approx\:10^{0.176}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Check
. . \(\displaystyle \begin{array}{cccc}10^{0.1760} & = & 1.499684836 \\ \; \\ \; \\10^{0.1761} & = & 1.500030190\end{array}\) . . . not bad!
Hello, NEHA!
Write \(\displaystyle \frac{3}{2}\) as a power with a base of 10.
We have: \(\displaystyle \:10^x\;=\;1.5\)
Take logs of both sides (base 10): \(\displaystyle \:\log\left(10^x\right) \;= \;\log(1.5)\)
Then we have: \(\displaystyle \:x\cdot\log(10) \;= \;\log(1.5)\)
Since \(\displaystyle \,\log(10)\,=\,1\), we have: \(\displaystyle \:x\;=\;\log(1.5) \;= \;0.176091259...\)
Therefore: \(\displaystyle \L\:\frac{3}{2} \:\approx\:10^{0.176}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Check
. . \(\displaystyle \begin{array}{cccc}10^{0.1760} & = & 1.499684836 \\ \; \\ \; \\10^{0.1761} & = & 1.500030190\end{array}\) . . . not bad!
Re: ~
THANKs
ok . so we can do the same way for this probem:
write 0.7165 as e raised to a power.
soroban said:Hello, NEHA!
Write \(\displaystyle \frac{3}{2}\) as a power with a base of 10.
We have: \(\displaystyle \:10^x\;=\;1.5\)
Take logs of both sides (base 10): \(\displaystyle \:\log\left(10^x\right) \;= \;\log(1.5)\)
Then we have: \(\displaystyle \:x\cdot\log(10) \;= \;\log(1.5)\)
Since \(\displaystyle \,\log(10)\,=\,1\), we have: \(\displaystyle \:x\;=\;\log(1.5) \;= \;0.176091259...\)
Therefore: \(\displaystyle \L\:\frac{3}{2} \:\approx\:10^{0.176}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Check
. . \(\displaystyle \begin{array}{cccc}10^{0.1760} & = & 1.499684836 \\ \; \\ \; \\10^{0.1761} & = & 1.500030190\end{array}\) . . . not bad!
THANKs
ok . so we can do the same way for this probem:
write 0.7165 as e raised to a power.