NCEA Level 2 Algebra Exam 2014 (part 2)

Another one from last year’s Algebra exam: Solve $\large{\boldsymbol{5^x\times 2^{-2x}=15}}$ .

You can find a copy of the exam question paper on the NZQA website here.

Question Three (b): Solve for $\boldsymbol{x}$ : $\boldsymbol{5^x\times 2^{-2x}=15}$ .

Knowing how to use the Exponent and Logarithm rules is vital to this one. In particular:

$\large{\boldsymbol{a^{-m}=\frac{1}{a^m}\;\text{ and }\; (a^m)^n=a^{mn}}}$

See if you can follow all the steps of my solution below:

$\large{\boldsymbol{5^x\times 2^{-2x}=15}} \\ \large{\boldsymbol{5^x\times \frac{1}{2^{2x}}=15}} \\ \large{\boldsymbol{\frac{5^x}{2^{2x}}=15}} \\ \large{\boldsymbol{\left(\frac{5}{2^2}\right)^x=15}} \\ \large{\boldsymbol{\log{\left(\frac{5}{4}\right)^x}=\log{15}}} \\ \large{\boldsymbol{x\log{\left(\frac{5}{4}\right)}=\log{15}}} \\ \large{\boldsymbol{x=\frac{\log{15}}{\log{\left(\frac{5}{4}\right)}}}}$

A common mistake would be to try to multiply the $\boldsymbol{5^x}$ and the $\boldsymbol{2^{-2x}}$ together directly by misunderstanding the exponent rule:

$\boldsymbol{a^m\times a^n=a^{m+n}}$

Notice that the base $\boldsymbol{a}$ is always the same in the rule above. We don’t have that in this case. If we had $\boldsymbol{5^x\times 5^{-2x}}$ , for example, then we could use that rule. But we don’t, so we must look elsewhere.

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