How do you solve for X in this equation?
Do you have any sources where I can read more about it?
3^x - 3^(x-1) = 3^5
Do you have any sources where I can read more about it?
3^x - 3^(x-1) = 3^5
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Thanks!kmittal82 wrote:Hmm, here's how I approached it, but got stuck at the end
3^(x-1) = (3^x)/3
3^x - 3^(x-1)
= 3^x - (3^x/3) = 3^5
Multiply both sides by 3
=> 3^(x+1) - 3^x = 3^6
Now, 3^(x+1 ) = 3^x * 3
Factoring out 3^x
3^x(3 - 1) = 3^6
=> 3^x = 3^6 x 0.5
This gives a non-integer value for x
Could you please give the OA and the source of the question?
But aren't logarithms outside the scope of the GMAT ?Maciek wrote:Hi Andre!
Search the forum for information on solving exponential equations. Look for expert replies.
you can check also this link:
https://www.purplemath.com/modules/solvexpo.htm
let us calculate it:
3^x - 3^(x-1) = 3^5
3^x - (1/3)*3^x = 3^5
3^x(1 - 1/3) = 3^5
2/3 * 3^x = 3 ^5
2 * 3^x = 3^6
3^x = (3^6)/2
log3 (3^x) = log3 ((3^6)/2)
x = log3 (3^6) - log3 (2)
x = 6 - log3 (2)
what are the answer choices?
----
log3 (2) = log10 (2)/log10 (3)
x = 6 - log10 (2)/log10 (3)
x = 6 - 0.3/0.48
x = 6 - 0.625
x = 5.375
hope it helps!
Best,
Maciek
@ andre.heggli - Going forward, can you please write the source when you post the question? Some of us are saving the GMAT Prep Test for later and solving a question from the test gives away a potential question from the test. Thanks!andre.heggli wrote:Thanks!kmittal82 wrote:Hmm, here's how I approached it, but got stuck at the end
3^(x-1) = (3^x)/3
3^x - 3^(x-1)
= 3^x - (3^x/3) = 3^5
Multiply both sides by 3
=> 3^(x+1) - 3^x = 3^6
Now, 3^(x+1 ) = 3^x * 3
Factoring out 3^x
3^x(3 - 1) = 3^6
=> 3^x = 3^6 x 0.5
This gives a non-integer value for x
Could you please give the OA and the source of the question?
The source is the GMAT Prep exams.
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