Sequence and Logarithmic

1.
log b - log a = d
log (b/a) = d
b/a = 10^d ...... [1]

Let r be the common ratio.
r = √(b^3) / √(a^3)
r = √[(b/a)^3] ...... [2]

Put [1] into [2] :
r = √[(10^d)^3]
r = (10^d)√(10^d)

The common ratio = (10^d)√(10^d)


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2.
(9^x) + (1/9^x) = 8
(3^x)^2 + [(1/3)^x]^2 = 8
(3^x)^2 + 2*(3^x)[(1/3)^x] + [(1/3)^x]^2 = 8 + 2*(3^x)[(1/3)^x]
[(3^x) + (1/3^x)]^2 = 10
(3^x) + (1/3^x) = √10 or (3^x) + (1/3^x) = -√10


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3.
log b / log a = log c / log b
(log b)² = log a * log c ...... [1]

log c / log b = log a / log c
(log c)² = log a * log b ...... [2]

log b / log a = log a / log c
(log a)² = log b * log c ...... [3]

[3]/[1] :
(log a)² / (log b)² = (log b * log c) / (log a * log c)
(log a)² / (log b)² = (log b) / (log a)
(log a)³ = (log b)³
log a = log b
a = b

[1]/[2] :
(log b)² / (log c)² = (log a * log c) / (log a * log b)
(log b)² / (log c)² = (log c) / (log b)
(log b)³ = (log c)³
log b = log c
b = c

Since a = b = c, thus
a : b : c = 1 : 1 : 1


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4.
3^(2x + 1) = 5^(x - 1)
log[3^(2x + 1)] = log[5^(x - 1)]
(2x + 1) log 3 = (x - 1) log 5
2x log 3 + log 3 = x log 5 - log 5
x log 5 - 2x log 3 = log 5 + log 3
x(log 5 - 2 log 3) = log 5 + log 3
x = (log 5 + log 3)/(log 5 - 2 log 3)