SOLUTION: I am having a few issues if someone could help. The first problem is
3 with exponent of x+4 = 7 with exponent of 2x-1
Through all my notes I am finding nothing like this, my
Algebra.Com
Question 1042042: I am having a few issues if someone could help. The first problem is
3 with exponent of x+4 = 7 with exponent of 2x-1
Through all my notes I am finding nothing like this, my exmples are where I would take 3 as my base for 2nd equation
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
3 with exponent of x+4 = 7 with exponent of 2x-1
----------------
3^(x+4) = 7^(2x-1)
---------
Since the variable is in the exponent, take the log::
(x+4)*log(3) = (2x-1)*log(7)
-------------------------------------
Now, get the x terms together::
x(log3-2*log(7)) = log(7)-4log(3)
x(-1.213) = -1.063
x = 0.877
-------------
Cheers,
Stan H.
--------------
RELATED QUESTIONS
I am having issues with
e with exponent of 2x minus e exponent x minus 6... (answered by stanbon)
I am soon to be going back to college and its been awhile since I've had math to do. I... (answered by rapaljer)
I am having issues with a word problem. Hoping someone could help me please.
A new... (answered by only4christ2)
Oh dear God please help me!!! We are studying Graphs of Exponential and Logarithmic... (answered by jim_thompson5910,scott8148)
I have the following problem. It has exponents so I will try to explain where they are.
(answered by solver91311)
hi i really need help with these problems I would greatly appreciate it if you could... (answered by jorel1380)
I have tried every way i can think of to simplify this type of equation and have looked... (answered by suresh)
I am having some issues with logarithm equations and need some help.
first problem: 2 (answered by lwsshak3)