Finding variable value
$begingroup$
$a$,$b$ and $c$ are real numbers each bigger than 1 such that
$$frac{2}{3}log_b{a} + frac{3}{5}log_c{b} + frac{5}{2}log_a{c} = 3$$
If the value of $b$ is 9 what must be the value of $a$?
I tried putting in the value of $b$ as 9 and then used the base change formula but couldn’t come up with a solution. Can someone help me out with the solution?
algebra-precalculus functions logarithms
edited Jan 15 at 6:48
gerw
20.1k11334
asked Jan 14 at 23:20
user601297user601297
41719
$endgroup$
add a comment |
$begingroup$
$a$,$b$ and $c$ are real numbers each bigger than 1 such that
$$frac{2}{3}log_b{a} + frac{3}{5}log_c{b} + frac{5}{2}log_a{c} = 3$$
If the value of $b$ is 9 what must be the value of $a$?
I tried putting in the value of $b$ as 9 and then used the base change formula but couldn’t come up with a solution. Can someone help me out with the solution?
algebra-precalculus functions logarithms
edited Jan 15 at 6:48
gerw
20.1k11334
asked Jan 14 at 23:20
user601297user601297
41719
$endgroup$
add a comment |
$begingroup$
$a$,$b$ and $c$ are real numbers each bigger than 1 such that
$$frac{2}{3}log_b{a} + frac{3}{5}log_c{b} + frac{5}{2}log_a{c} = 3$$
If the value of $b$ is 9 what must be the value of $a$?
I tried putting in the value of $b$ as 9 and then used the base change formula but couldn’t come up with a solution. Can someone help me out with the solution?
algebra-precalculus functions logarithms
edited Jan 15 at 6:48
gerw
20.1k11334
asked Jan 14 at 23:20
user601297user601297
41719
$endgroup$
$a$,$b$ and $c$ are real numbers each bigger than 1 such that
$$frac{2}{3}log_b{a} + frac{3}{5}log_c{b} + frac{5}{2}log_a{c} = 3$$
If the value of $b$ is 9 what must be the value of $a$?
I tried putting in the value of $b$ as 9 and then used the base change formula but couldn’t come up with a solution. Can someone help me out with the solution?
algebra-precalculus functions logarithms
algebra-precalculus functions logarithms
edited Jan 15 at 6:48
gerw
20.1k11334
asked Jan 14 at 23:20
user601297user601297
41719
edited Jan 15 at 6:48
gerw
20.1k11334
asked Jan 14 at 23:20
user601297user601297
41719
edited Jan 15 at 6:48
gerw
20.1k11334
edited Jan 15 at 6:48
gerw
20.1k11334
edited Jan 15 at 6:48
gerw
20.1k11334
20.1k11334
asked Jan 14 at 23:20
user601297user601297
41719
asked Jan 14 at 23:20
user601297user601297
41719
asked Jan 14 at 23:20
user601297user601297
41719
41719
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
$log_b a = frac {log a}{log b}$
$frac {2log a}{3log b} + frac {3log b}{5log c} + frac {5log c}{2log a} = 3$
let $alpha, beta, gamma = 2log a, 3log b, 5log c$ respectively
The AM-GM inequality.
$frac 13 (frac {alpha}{beta} + frac {beta}{gamma} + frac {gamma}{alpha}) ge (frac {alpha}{beta} frac {beta}{gamma} frac {gamma}{alpha})^frac 13$
$(frac {alpha}{beta} frac {beta}{gamma} frac {gamma}{alpha})^frac 13 = 1$
If
$(frac {alpha}{beta} + frac {beta}{gamma} + frac {gamma}{alpha}) = 3$
We can conclude:
$frac {alpha}{beta} = frac {beta}{gamma} = frac {gamma}{alpha} = 1$
$2log a = 3log b = 5log c\
a^2 = b^3 = c^5$
$b = 9 = 3^2\
a^2= b^3 = 3^6\
a = 3^3\
c = 3^{frac 65}$
answered Jan 14 at 23:42
Doug MDoug M
45.4k31954
$endgroup$
$begingroup$
Amazing solution, how did it strike you to use the A.M. G.M. Inequality? Thanks a lot
$endgroup$
– user601297
Jan 14 at 23:46
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
$log_b a = frac {log a}{log b}$
$frac {2log a}{3log b} + frac {3log b}{5log c} + frac {5log c}{2log a} = 3$
let $alpha, beta, gamma = 2log a, 3log b, 5log c$ respectively
The AM-GM inequality.
$frac 13 (frac {alpha}{beta} + frac {beta}{gamma} + frac {gamma}{alpha}) ge (frac {alpha}{beta} frac {beta}{gamma} frac {gamma}{alpha})^frac 13$
$(frac {alpha}{beta} frac {beta}{gamma} frac {gamma}{alpha})^frac 13 = 1$
If
$(frac {alpha}{beta} + frac {beta}{gamma} + frac {gamma}{alpha}) = 3$
We can conclude:
$frac {alpha}{beta} = frac {beta}{gamma} = frac {gamma}{alpha} = 1$
$2log a = 3log b = 5log c\
a^2 = b^3 = c^5$
$b = 9 = 3^2\
a^2= b^3 = 3^6\
a = 3^3\
c = 3^{frac 65}$
answered Jan 14 at 23:42
Doug MDoug M
45.4k31954
$endgroup$
$begingroup$
Amazing solution, how did it strike you to use the A.M. G.M. Inequality? Thanks a lot
$endgroup$
– user601297
Jan 14 at 23:46
add a comment |
$begingroup$
$log_b a = frac {log a}{log b}$
$frac {2log a}{3log b} + frac {3log b}{5log c} + frac {5log c}{2log a} = 3$
let $alpha, beta, gamma = 2log a, 3log b, 5log c$ respectively
The AM-GM inequality.
$frac 13 (frac {alpha}{beta} + frac {beta}{gamma} + frac {gamma}{alpha}) ge (frac {alpha}{beta} frac {beta}{gamma} frac {gamma}{alpha})^frac 13$
$(frac {alpha}{beta} frac {beta}{gamma} frac {gamma}{alpha})^frac 13 = 1$
If
$(frac {alpha}{beta} + frac {beta}{gamma} + frac {gamma}{alpha}) = 3$
We can conclude:
$frac {alpha}{beta} = frac {beta}{gamma} = frac {gamma}{alpha} = 1$
$2log a = 3log b = 5log c\
a^2 = b^3 = c^5$
$b = 9 = 3^2\
a^2= b^3 = 3^6\
a = 3^3\
c = 3^{frac 65}$
answered Jan 14 at 23:42
Doug MDoug M
45.4k31954
$endgroup$
$begingroup$
Amazing solution, how did it strike you to use the A.M. G.M. Inequality? Thanks a lot
$endgroup$
– user601297
Jan 14 at 23:46
add a comment |
$begingroup$
$log_b a = frac {log a}{log b}$
$frac {2log a}{3log b} + frac {3log b}{5log c} + frac {5log c}{2log a} = 3$
let $alpha, beta, gamma = 2log a, 3log b, 5log c$ respectively
The AM-GM inequality.
$frac 13 (frac {alpha}{beta} + frac {beta}{gamma} + frac {gamma}{alpha}) ge (frac {alpha}{beta} frac {beta}{gamma} frac {gamma}{alpha})^frac 13$
$(frac {alpha}{beta} frac {beta}{gamma} frac {gamma}{alpha})^frac 13 = 1$
If
$(frac {alpha}{beta} + frac {beta}{gamma} + frac {gamma}{alpha}) = 3$
We can conclude:
$frac {alpha}{beta} = frac {beta}{gamma} = frac {gamma}{alpha} = 1$
$2log a = 3log b = 5log c\
a^2 = b^3 = c^5$
$b = 9 = 3^2\
a^2= b^3 = 3^6\
a = 3^3\
c = 3^{frac 65}$
answered Jan 14 at 23:42
Doug MDoug M
45.4k31954
$endgroup$
$log_b a = frac {log a}{log b}$
$frac {2log a}{3log b} + frac {3log b}{5log c} + frac {5log c}{2log a} = 3$
let $alpha, beta, gamma = 2log a, 3log b, 5log c$ respectively
The AM-GM inequality.
$frac 13 (frac {alpha}{beta} + frac {beta}{gamma} + frac {gamma}{alpha}) ge (frac {alpha}{beta} frac {beta}{gamma} frac {gamma}{alpha})^frac 13$
$(frac {alpha}{beta} frac {beta}{gamma} frac {gamma}{alpha})^frac 13 = 1$
If
$(frac {alpha}{beta} + frac {beta}{gamma} + frac {gamma}{alpha}) = 3$
We can conclude:
$frac {alpha}{beta} = frac {beta}{gamma} = frac {gamma}{alpha} = 1$
$2log a = 3log b = 5log c\
a^2 = b^3 = c^5$
$b = 9 = 3^2\
a^2= b^3 = 3^6\
a = 3^3\
c = 3^{frac 65}$
answered Jan 14 at 23:42
Doug MDoug M
45.4k31954
answered Jan 14 at 23:42
Doug MDoug M
45.4k31954
answered Jan 14 at 23:42
Doug MDoug M
45.4k31954
answered Jan 14 at 23:42
Doug MDoug M
45.4k31954
45.4k31954
$begingroup$
Amazing solution, how did it strike you to use the A.M. G.M. Inequality? Thanks a lot
$endgroup$
– user601297
Jan 14 at 23:46
add a comment |
$begingroup$
Amazing solution, how did it strike you to use the A.M. G.M. Inequality? Thanks a lot
$endgroup$
– user601297
Jan 14 at 23:46
$begingroup$
Amazing solution, how did it strike you to use the A.M. G.M. Inequality? Thanks a lot
$endgroup$
– user601297
Jan 14 at 23:46
$begingroup$
Amazing solution, how did it strike you to use the A.M. G.M. Inequality? Thanks a lot
$endgroup$
– user601297
Jan 14 at 23:46
add a comment |
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