How do you solve #4^(2x+1)= 1024#?

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Answer 1 · 2018-02-28T23:01:48

Use natural logarithm on both sides:

#ln(4^(2x+1))= ln(1024)#

Use the property of logarithms that allows one to move the exponent to the outside as a factor:

#(2x+1)ln(4)= ln(1024)#

Divide both sides by #ln(4)#:

#2x+1= ln(1024)/ln(4)#

Subtract 1 from both sides:

#2x= ln(1024)/ln(4)-1#

Divide both sides by 2:

#x= ln(1024)/(2ln(4))-1/2#

Use a calculator:

#x = 2#

Answer 2 · 2018-02-28T23:04:08

Use a logarithm

I prefer natural log, ln, although you could use base 10 common log also.

So, following the rule that you can do whatever you want to an equation as long as you do the same thing to both sides:

#ln 4^{2x+1} = ln 1024#

Then, following logarithm rules, ln #x^n# = n ln x

So, #(2x+1) ln 4 = ln 1024#

At this point, you can start to isolate x. Divide both sides by ln 4.

#2x + 1 = {ln 1024}/{ln 4}#

Sub 1 from both sides and divide by 2. Of course you can evaluate your partial answer any time. Example: #{ln 1024}/{ln 4}#= 5

This gives #x = {{ln 1024}/{ln 4}-1}/2->x=2#

Check your answer: #4^{2*2+1}->4^5=1024#