Thread: logarithmic question

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need some help urgently

Solve the equation.
(16/9)^{x-1} = (3/4)^{x-1}

Originally Posted by math321

need some help urgently

Solve the equation.
(16/9)^{x-1} = (3/4)^{x-1}

log both sides so that you can bring down the powers.





now continue (note that and are constants). where can you go from there?

So am i finding the log 16/9 and log 3/4 and multiply it by x-1 and solve for x

Originally Posted by math321

So am i finding the log 16/9 and log 3/4 and multiply it by x-1 and solve for x

yes, but just leave them as log (16/9) and log (3/4), no need to evaluate them. so you're on the right track. multiplying out and getting all x's on one side would be the way to proceed.

can u verify this anser for me please

(16/9)^(x-1) = (9/16)^-(x-1) = (9/16)^(1-x) = ((3/4)^2)^(1-x) = (3/4)^(2(1-x))

(3/4)^(2(1-x)) = (3/4)^(x-1)

2(1-x) = (x-1)
2-2x = x-1
3 = 3x
x = 1

Originally Posted by math321

can u verify this anser for me please

(16/9)^(x-1) = (9/16)^-(x-1) = (9/16)^(1-x) = ((3/4)^2)^(1-x) = (3/4)^(2(1-x))

(3/4)^(2(1-x)) = (3/4)^(x-1)

2(1-x) = (x-1)
2-2x = x-1
3 = 3x
x = 1

yes, that works!