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Evaluate without using a calculator: $(\log_{ 7} 49)^5 + \log_{ 5} \: \biggl(\dfrac{1}{125}\biggr) - 13\:\log_{ 9} 1$
in Grade 12 Maths by Platinum (114k points) | 24 views

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Take note 1/125 = 5^-3, 49= 7^2, x represents the exponents/logarithms of the numbers.

(7^x)^5 + (5^x) - 13(9^x)

= 2^5 + log 5^-3 - 13( 9^x)

= 32 + (-3) -13(0)

=29
by Diamond (85.5k points)

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