Can someone solve for x please. I don't understand it.
2^x . 7^x = log 14
My answer : 14^x = log 14
= log14÷log 14 = x
X=1
Is this right????
yes, it is right.
a^x * b^x= (ab)^x
2^2x+1 - (9)2^x = -4
Can someone please solve for x ?
My answer is x= 2 , is this correct?? Please show your steps
Hi bob
For the first log problem , it was in a test and my answer was wrong , could it be a marking error?
is the problem this
2^(2x+1) -9*2^2x = -4
or this
2^(2x) +1 -9*2^2x = -4
Sorry it wasn't clear the problem was this 2^(2x+1) -9*2^2x = -4
yes, it is correct
#1 answer is incorrect
2^x * 6^x = log 14
14^x = log 14
now take logs of both sides, and use log rules
x log14 = log(log14)
x = log(log14)/log14 = appr .05168
it is easy to show that x = 1 is not a solution to the question
since 14^1 ≠ log14
#2
2^2x+1 - (9)2^x = -4
the way you typed it, x = 2 is definitely not a solution
LS = 4(2)+1-9(4) = -27 ≠ -4
if you meant:
2^(2x+1) - (9)2^x = -4 , then
2(2^(2x) - 9(2^x) + 4 = 0
2 (2^x)^2 - 9(2^x) + 4 = 0
let 2^x = y , then we have
2y^2 - 9y + 4 = 0
(y - 4)(2y - 1) = 0
y = 4 or y = 1/2
then 2^x = 4 -----> x = 2
or 2^x = 1/2 ------ x = -1
Thank you Reiny
Your answer is very clear and shows what I did wrong. Keep up the good work.
To solve for x in the equation 2^x * 7^x = log 14, you first need to simplify the equation.
Start by using the property of logarithms, log(a*b) = log(a) + log(b), to rewrite the equation:
log(2^x) + log(7^x) = log 14
Now, apply the logarithmic property log(a^n) = n*log(a):
x * log(2) + x * log(7) = log 14
Next, factor out the common factor of x:
x * (log(2) + log(7)) = log 14
Finally, divide both sides of the equation by (log(2) + log(7)) to solve for x:
x = log 14 / (log(2) + log(7))
Now, let's evaluate this expression:
x ≈ log 14 / (log 2 + log 7)
Using a calculator, you can find the values for log 14, log 2, and log 7. Substitute those values into the equation to get the approximate value for x.