# solve this 5 + log base 2 ^3?

Okay so 2^z if z= 5 + log(base2) 3 solve for z

Need help asap

### 3 Answers

- la consoleLv 72 months ago
First possibility:

= 2^(z) → where: z= 5 + Log_2(3) = Log[2](96)

= 2^{5 + Log_2(3)]

= 2^(5) * 2^[Log_2(3)]

= 2^(5) * 2^[Ln(3)/Ln(2)]

= 2^(5) * {2^[Ln(3)]}^[Ln(2)]

Second possibility:

z= 5 + Log[2](3) → recall: Log{a}(x) = Ln(x) / Ln(a) ← where a is the base

z= 5 + [Ln(3)/Ln(2)]

z= [5.Ln(2) + Ln(3)]/Ln(2)

z= [Ln(2⁵) + Ln(3)]/Ln(2)

z= [Ln(32) + Ln(3)]/Ln(2)

z= [Ln(32 * 3)]/Ln(2)

z= Ln(96)/Ln(2)

z= Log[2](96)

= 2^(z) → where: z= 5 + Log_2(3) = Log[2](96)

= 2^[Log[2](96)]

- ?Lv 72 months ago
z = 5 + log₂ 3

log 3

log₂ 3 = --------- ≈ 1.58

log 2

z = 5 + 1.58 = 6.58 ...............ANS

- rotchmLv 72 months ago
2^(5 + L3) = 2^5 * 2^(L3) = 2^5 * 3.

Can you justify each of these steps?

Done!

[PS: Do you really want to solve for z, or to solve for 2^z ?].