Answer: The answer is 80. Step-by-step explanation: 10ˣ = 64 ⇔ log₁₀64 = x log₁₀8² = x ⇒ 2×log₁₀8 = x ⇒ log₁₀8 = x/2 ..1 To find 10^(x/2)+1 substitute x/2 = log₁₀8 10^( log₁₀8 + 1) = y 10^(log₁₀8 + log₁₀10) = y (note : logₓx = 1 and logₓa + logₓb = logₓab) 10^(log₁₀80) = y log₁₀10^(log₁₀80) = log₁₀y (log₁₀80)×(log₁₀10) = log₁₀y (note : logₓ(a^logₙb) = (logₙb) × ( logₓa) ) log₁₀80 × 1 = log₁₀y 80 = y y = 10^((x/2)+1) = 80 Hope it helps. Mark as brainliest if it was helpful. Thank you. Reply

Answer:The answer is 80.

Step-by-step explanation:10ˣ = 64 ⇔ log₁₀64 = x

log₁₀8² = x ⇒ 2×log₁₀8 = x ⇒ log₁₀8 = x/2 ..1

To find 10^(x/2)+1

substitute x/2 = log₁₀8

10^( log₁₀8 + 1) = y

10^(log₁₀8 + log₁₀10) = y

(note : logₓx = 1 and logₓa + logₓb = logₓab)10^(log₁₀80) = y

log₁₀10^(log₁₀80) = log₁₀y

(log₁₀80)×(log₁₀10) = log₁₀y

(note : logₓ(a^logₙb) = (logₙb) × ( logₓa) )log₁₀80 × 1 = log₁₀y

80 = y

y = 10^((x/2)+1) = 80

Hope it helps. Mark as brainliest if it was helpful. Thank you.