Algebraic Manipulation and Logarithms

Dr Fua

New member
Joined
Dec 2, 2015
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16
today, i have encountered a not so usual (for me at least) exponential equations. Now, i could solve it easily if i manipulated the equation like a common algebraic one, but the problems is... i am not quite sure if i am allowed to do so; since, the properties of logarithms is very limited as it is...

db8a53bacc.png





and here is what i tried to do....

12334572_10204558619197989_1216306811_o.jpg
 
today, i have encountered a not so usual (for me at least) exponential equations. Now, i could solve it easily if i manipulated the equation like a common algebraic one, but the problems is... i am not quite sure if i am allowed to do so; since, the properties of logarithms is very limited as it is...

db8a53bacc.png





and here is what i tried to do....

Your method and answer are correct..

(3x-1)*ln(2) = (5x+8)*ln(7)

x * 3*ln(2) - x * 5 * ln(7) = 8*ln(7) + ln(2)

x = [8*ln(7) + ln(2)] / [3*ln(2) - 5 * ln(7)]
 
What's up, Doc?
Perhaps looking at it "in detail" will help out "seeing" how it works:

2^(3x - 1) = 2^(3x)*2^(-1) = (2^3)^x / 2 [1]

7^(5x + 8) = 7^(5x)*7^8 = (7^5)^x * 7^8 [2]

[1] = [2]:
(2^3)^x / 2 = (7^5)^x * 7^8
(2^3)^x = (7^5)^x * 7^8 * 2
(2^3)^x / (7^5)^x = 7^8 * 2

let u = 7^8 * 2; so:
[(2^3) / (7^5)]^x = u

let v = (2^3) / (7^5); so:
v^x = u
x = log(u) / log(v)
x = -2.12551585....

Hope that helps...

very helpful! thanks!
 
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