Middle School Log question
$$left(frac 43right)^x=frac{8sqrt3}{9}$$
I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.
logarithms
edited 1 hour ago
Rócherz
2,7362721
asked 1 hour ago
Toylatte
113
add a comment |
$$left(frac 43right)^x=frac{8sqrt3}{9}$$
I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.
logarithms
edited 1 hour ago
Rócherz
2,7362721
asked 1 hour ago
Toylatte
113
3
If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
1 hour ago
add a comment |
$$left(frac 43right)^x=frac{8sqrt3}{9}$$
I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.
logarithms
edited 1 hour ago
Rócherz
2,7362721
asked 1 hour ago
Toylatte
113
$$left(frac 43right)^x=frac{8sqrt3}{9}$$
I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.
logarithms
logarithms
edited 1 hour ago
Rócherz
2,7362721
asked 1 hour ago
Toylatte
113
edited 1 hour ago
Rócherz
2,7362721
asked 1 hour ago
Toylatte
113
edited 1 hour ago
Rócherz
2,7362721
edited 1 hour ago
Rócherz
2,7362721
edited 1 hour ago
Rócherz
2,7362721
2,7362721
asked 1 hour ago
Toylatte
113
asked 1 hour ago
Toylatte
113
asked 1 hour ago
Toylatte
113
113
3
If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
1 hour ago
add a comment |
3
If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
1 hour ago
3
3
If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
1 hour ago
If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
1 hour ago
add a comment |
2 Answers
2
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oldest
votes
I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
$$
left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
$$
Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!
Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.
answered 1 hour ago
ImNotTheGuy
3825
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ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Thanks it helped
– Toylatte
1 hour ago
@Toylatte You are welcome!
– ImNotTheGuy
1 hour ago
add a comment |
$$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$
Note that the function $f(x)=left( frac43right)^x$ is an increasing function.
Hence $x=1.5$.
Alternatively, taking logarithm on both sides.
answered 1 hour ago
Siong Thye Goh
98.8k1464116
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
$$
left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
$$
Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!
Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.
answered 1 hour ago
ImNotTheGuy
3825
New contributor
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thanks it helped
– Toylatte
1 hour ago
@Toylatte You are welcome!
– ImNotTheGuy
1 hour ago
add a comment |
I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
$$
left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
$$
Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!
Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.
answered 1 hour ago
ImNotTheGuy
3825
New contributor
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thanks it helped
– Toylatte
1 hour ago
@Toylatte You are welcome!
– ImNotTheGuy
1 hour ago
add a comment |
I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
$$
left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
$$
Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!
Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.
answered 1 hour ago
ImNotTheGuy
3825
New contributor
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
$$
left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
$$
Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!
Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.
answered 1 hour ago
ImNotTheGuy
3825
New contributor
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
ImNotTheGuy
3825
New contributor
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
ImNotTheGuy
3825
answered 1 hour ago
ImNotTheGuy
3825
3825
New contributor
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thanks it helped
– Toylatte
1 hour ago
@Toylatte You are welcome!
– ImNotTheGuy
1 hour ago
add a comment |
Thanks it helped
– Toylatte
1 hour ago
@Toylatte You are welcome!
– ImNotTheGuy
1 hour ago
Thanks it helped
– Toylatte
1 hour ago
Thanks it helped
– Toylatte
1 hour ago
@Toylatte You are welcome!
– ImNotTheGuy
1 hour ago
@Toylatte You are welcome!
– ImNotTheGuy
1 hour ago
add a comment |
$$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$
Note that the function $f(x)=left( frac43right)^x$ is an increasing function.
Hence $x=1.5$.
Alternatively, taking logarithm on both sides.
answered 1 hour ago
Siong Thye Goh
98.8k1464116
add a comment |
$$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$
Note that the function $f(x)=left( frac43right)^x$ is an increasing function.
Hence $x=1.5$.
Alternatively, taking logarithm on both sides.
answered 1 hour ago
Siong Thye Goh
98.8k1464116
add a comment |
$$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$
Note that the function $f(x)=left( frac43right)^x$ is an increasing function.
Hence $x=1.5$.
Alternatively, taking logarithm on both sides.
answered 1 hour ago
Siong Thye Goh
98.8k1464116
$$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$
Note that the function $f(x)=left( frac43right)^x$ is an increasing function.
Hence $x=1.5$.
Alternatively, taking logarithm on both sides.
answered 1 hour ago
Siong Thye Goh
98.8k1464116
answered 1 hour ago
Siong Thye Goh
98.8k1464116
answered 1 hour ago
Siong Thye Goh
98.8k1464116
answered 1 hour ago
Siong Thye Goh
98.8k1464116
98.8k1464116
add a comment |
add a comment |
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If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
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