# Maths , logarithms

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????

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1. yes, it is right.

a^x * b^x= (ab)^x

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bobpursley
2. 2^2x+1 - (9)2^x = -4

Can someone please solve for x ?

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3. Hi bob
For the first log problem , it was in a test and my answer was wrong , could it be a marking error?

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4. is the problem this
2^(2x+1) -9*2^2x = -4
or this
2^(2x) +1 -9*2^2x = -4

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bobpursley
5. Sorry it wasn't clear the problem was this 2^(2x+1) -9*2^2x = -4

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6. yes, it is correct

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bobpursley

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

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8. Thank you Reiny
Your answer is very clear and shows what I did wrong. Keep up the good work.

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