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Logarithim Help.

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hobbes
Posts: 898 - Karma: 17
 If "Log a B = c^4" and "Log b A = 5/c^3" prove that c= 1/5


lower case letters are suppsoed to be subscripts and the "^" sign is exponet and the "/" is a fraction sign.

Any help would be great.
Apr 16 2007 8:39PM
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Gavajabra2
Posts: 1259 - Karma: 13
A x c^4 = B, try figuring it out that way?
Apr 16 2007 8:42PM
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quixotic
Posts: 6937 - Karma: 2,978
thats pre calc right?
Apr 16 2007 8:47PM
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hobbes
Posts: 898 - Karma: 17
it's definitley Advanced Math 12, which is a step up from grade 12 adcademic math but I am taking it in grade 11 to take pre- cal next year.
Apr 16 2007 8:49PM
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user098123
Posts: 6072 - Karma: 18
If "Log a B = c^4" and "Log b A = 5/c^3" prove that c= 1/5

Use antilogs. If Log a B = C^4 then a^(C^4) = B, and for Log b A = 5/(C^3), b^(5/(C^3)) = A, solve from there.
Apr 16 2007 9:00PM
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mmmmm...Pie
Posts: 673 - Karma: 15
Log a B = C^4 --> B = A * C^4

Log b A = 5/C^3 --> A = B * 5/C^3

Sub in "B*5/C^3" for A --> B = B * 5/C^3 * C^4

C's cancel --> B = B * 5 * C

Divide both sides by 5B (B's cancel) --> 1/5 = C
Apr 16 2007 9:01PM
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squeakywaffle
Posts: 1800 - Karma: 13
Since log a b = c^4, a^(c^4) = b. So, a = b^(1/c^4). So, bringing the other equation into the mix, we get:

b^(1/c^4) = b^(5/c^3)

So, 1/c^4 = 5/c^3 and 1/5 = c^4/c^3. Cancel on the right side and you get:

1/5 = c.

thankyouverymuch
Apr 16 2007 9:03PM
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hobbes
Posts: 898 - Karma: 17
Thank you very much everyone who helped out.
Apr 16 2007 9:14PM
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