Logarithms/exponential equations

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?

Reply 1

lazy_fish

1

Original post by Mysticmeg

...

x^a = x^b \Rightarrow a = b

4 = 2^2

(edited 9 years ago)

Reply 2

bethanoconnor

2

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) :smile:

Posted from TSR Mobile

Reply 3

davros

16

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?