Question 1158662
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Throughout the entire solution below, all logs are base 'a'.
Instead of writing log'a'(x) to mean "log base 'a' of x", I'm simply going to write "log(x)". 
Alternatively I could write log(a,x) to mean "log base 'a' of x", but again I'll just stick with log(x) for simplicity.


log(b^2) = c is given


We'll use the rule y*log(x) = log(x^y) to transform the exponent of '2' into an exponent of '5'


Multiply both sides by (5/2) and apply the log rule shown above
log(b^2) = c
(5/2)log(b^2) = (5/2)c
log( (b^2)^(5/2) ) = (5/2)c
log( b^(2*(5/2)) ) = (5/2)c
log( b^5 ) = (5/2)c
The exponents 2 and 5/2 multiply to 5.


We are able to replace log( b^5 ) with (5/2)c.


Next use the log rule log(xy) = log(x)+log(y)
log(7b^5) = log(7)+log(b^5)
log(7b^5) = log(7)+(5/2)c



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Therefore,
log(7b^5) + log(a^3) - log(7) = log(7)+(5/2)c + 3*log(a) - log(7)
log(7b^5) + log(a^3) - log(7) = log(7)+(5/2)c + 3*1 - log(7)
log(7b^5) + log(a^3) - log(7) = (5/2)c + 3 + log(7)-log(7)
log(7b^5) + log(a^3) - log(7) = <font color=red>(5/2)c + 3</font>


Again, all logs in this solution are base 'a'. 
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