# Logarithms/exponential equationsWatch

#1
I'm stuck on part of a question:

9) Solve these questions without using a calculator
(a) 6^x = 6^(2x+1)

(b) 4^(4x+6)=4^(7x-3)

(c) 2^(4x+6)=4^(7x-3)

I can do part (a):
6^x= (6^x)^2(6^1)
6y^2 - y
y(6y-1)
y=0, y=1/6
6^x = 1/6
so x=-1

But I can't figure out how to do part (b) or (c)
My attempt at (B):
(4^x)^4(4^6)= (4^x)^7(4^-3)
(4^6)Y^4=y^7(4^-3)
(3096)y^4=(1/64)y^7
this looks way too complicated.
Also how do you do part (c) because there are two different bases?
0
4 years ago
#2
(Original post by Mysticmeg)
...

1
4 years ago
#3
If the base numbers are the same then just let the powers equal each other and then solve for x!
It's actually way more simple than how you did part (a)

Posted from TSR Mobile
0
4 years ago
#4
(Original post by Mysticmeg)
I'm stuck on part of a question:

9) Solve these questions without using a calculator
(a) 6^x = 6^(2x+1)

(b) 4^(4x+6)=4^(7x-3)

(c) 2^(4x+6)=4^(7x-3)

I can do part (a):
6^x= (6^x)^2(6^1)
6y^2 - y
y(6y-1)
y=0, y=1/6
6^x = 1/6
so x=-1

But I can't figure out how to do part (b) or (c)
My attempt at (B):
(4^x)^4(4^6)= (4^x)^7(4^-3)
(4^6)Y^4=y^7(4^-3)
(3096)y^4=(1/64)y^7
this looks way too complicated.
Also how do you do part (c) because there are two different bases?
You're making heavy weather of this!

If then a = b so for part (a) you can write x = 2x + 1 and solve giving x = -1 as you have correctly found.

Now try the same method for part (b).
0
4 years ago
#5
(Original post by lazy_fish)

I was just a little curious as to what conditions should be placed on for the implication you mention to hold. Definitely, we have that ? Any other conditions?
0
#6
(Original post by lazy_fish)

So for (c) would I
4x + 6 = 7x - 3

square root the left side or square the left side?
0
4 years ago
#7
(Original post by Mysticmeg)
So for (c) would I
4x + 6 = 7x - 3

square root the left side or square the left side?
Don't square root either side!!

Express both sides as powers of 2 and then follow the same process as for parts (a) and (b).
0
4 years ago
#8
Rewrite 4 as , and then remember that . Once the bases are the same, you can simply equate the powers then solve for
0
#9
Thanks
0
4 years ago
#10
(Original post by lazy_fish)

I owed you a rep...
0
X

new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Cardiff Metropolitan University
Sat, 27 Apr '19
• University of East Anglia
Could you inspire the next generation? Find out more about becoming a Primary teacher with UEA… Postgraduate
Sat, 27 Apr '19
• Anglia Ruskin University
Health, Education, Medicine and Social Care; Arts, Humanities and Social Sciences; Business and Law; Science and Engineering Undergraduate
Sat, 27 Apr '19

### Poll

Join the discussion

#### Have you registered to vote?

Yes! (564)
37.65%
No - but I will (117)
7.81%
No - I don't want to (108)
7.21%
No - I can't vote (<18, not in UK, etc) (709)
47.33%