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how would i solve 19^(4x-15)=11?
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19^(4x-15)=11 take the natural logarithm of both sides of the equation to remove the variable from the exponent. ln(19^(4x-15))=ln(11) The lefthand side of the equation is equal to the exponent of the logarithm argument because the base of the logarithm equals the base of the argument. 4xln(19)-15ln(19)=ln(11) Since -15ln(19) does not contain the variable to solve for, move it to the righthand side of the equation by adding 15ln(19) to both sides. 4xln(19)=15ln(19)+ln(11) Divide each term in the equation by 4ln(19). x=(15ln(19)+ln(11))/(4ln(19)) x=3.9536
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