Question

no calculator is allowed on this question.
the function ( f(x) = 10^x ) is graphed on a semi - log plot where the ( y ) - axis is logarithmically scaled. it is then transformed to make the graph of ( g(f(x)) ) shown above.
image created using desmos (cc by - sa 4.0)
what is ( g(x) )?
select one answer
a ( g(x)=x + 3 )
b ( g(x)=3cdot10^x )
c ( g(x)=\frac{1}{3}x )
d ( g(x)=3x )

Explanation:

Step1: Analyze the semi - log plot

In a semi - log plot where the \(y\) - axis is logarithmically scaled, if \(f(x)=10^{x}\), then \(\log(y)\) (where \(y = g(f(x))\)) will have a linear relationship with \(x\) if \(g\) is a logarithmic function or a linear function of a logarithmic function. But let's first consider the form of \(g(f(x))\). Since \(f(x) = 10^{x}\), let's assume \(g(x)\) is a linear function first. Let's look at the point \((0,3)\) on the graph of \(y = g(f(x))\). When \(x = 0\), \(f(0)=10^{0}=1\). So \(g(f(0))=g(1)\). Wait, no, maybe we should think about the transformation. Wait, the graph of \(g(f(x))\) is a straight line. Let's suppose \(g(f(x))=mx + b\). When \(x = 0\), \(f(0)=10^{0}=1\), and from the graph, when \(x = 0\), \(y = 3\). So \(g(1)=3\). Let's check the options:

- Option A: \(g(x)=x + 3\). Then \(g(1)=1 + 3=4 eq3\), so A is wrong.
- Option B: \(g(x)=3\cdot10^{x}\). Then \(g(1)=3\times10^{1}=30 eq3\), so B is wrong.
- Option C: \(g(x)=\frac{1}{3}x\). Then \(g(1)=\frac{1}{3} eq3\), so C is wrong.
- Option D: \(g(x)=3x\). Then \(g(1)=3\times1 = 3\), which matches the point \((0,3)\) (since when \(x = 0\), \(f(0)=1\), \(g(f(0))=g(1)=3\)). Let's check the linearity. Since \(f(x)=10^{x}\), \(g(f(x)) = 3\times10^{x}\)? Wait, no, wait. Wait, maybe I made a mistake. Wait, the graph is of \(g(f(x))\). If \(f(x)=10^{x}\), and the graph is a straight line, in a semi - log plot (log - y axis), the equation of a straight line is \(\log(y)=mx + b\), but here the y - axis is log - scaled, so the actual \(y\) - value (before log - scaling) is \(y = 10^{mx + b}\). But the graph shown is a straight line in the semi - log plot, which means that \(g(f(x))\) is a linear function of \(x\) in the log - y scale. Wait, maybe a better approach: Let's assume that \(g(x)\) is a linear function. Let's suppose \(g(f(x))=ax + c\). When \(x = 0\), \(f(0)=1\), so \(g(1)=c\). From the graph, when \(x = 0\), \(y = 3\), so \(c = 3\)? No, wait, maybe I mixed up. Wait, the function \(f(x)=10^{x}\) is graphed on a semi - log plot (y - axis log - scaled), so the original plot of \(f(x)\) would be an exponential curve, but after applying \(g\), it becomes a straight line. A straight line in a semi - log plot (log - y) means that \(g(f(x))\) is a linear function of \(x\) if \(g\) is a logarithmic function, but if \(g\) is a linear function, let's see:

Wait, let's re - express. Let \(y = g(f(x))\). Since \(f(x)=10^{x}\), \(y = g(10^{x})\). The graph of \(y\) vs \(x\) is a straight line. Let's find the equation of the straight line. We know that when \(x = 0\), \(y = 3\). Let's assume the slope. Wait, maybe the key is that when \(x = 0\), \(f(0)=1\), so \(g(1)=3\). Let's check each option:

- For option D: \(g(x)=3x\), \(g(1)=3\times1 = 3\), which matches. Let's check another point. Suppose we take \(x = 1\), \(f(1)=10^{1}=10\), \(g(10)=3\times10 = 30\). If the line has a slope, let's see, from \(x = 0\), \(y = 3\) and if the slope is 3 (since \(g(x)=3x\)), then the equation of the line \(y = 3x\) (but \(y = g(f(x))\)). Wait, maybe I was wrong earlier. Wait, the function \(f(x)=10^{x}\), so \(g(f(x))\) is the function we see. If \(g(x)=3x\), then \(g(f(x))=3\times10^{x}\), which is an exponential function, but in a semi - log plot (log - y axis), \(\log(y)=\log(3)+x\), which is a straight line with slope 1 and y - intercept \(\log(3)\). But the graph we have is a straight line with slope, let's see, from the point \((0,3)\), if we consider the x - axis as linear and y - axis as log - scaled, the equation of the line in log - y form is \(\log(y)=mx + b\). When \(x = 0\), \(\log(3)=b\) (since \(y = 3\) when \(x = 0\)). If \(g(x)=3x\), then \(y = g(f(x))=3\times10^{x}\), so \(\log(y)=\log(3)+x\), which is a straight line with slope 1 and y - intercept \(\log(3)\), which is consistent with a straight line in a semi - log plot.

Wait, but let's check the other options again. Option A: \(g(x)=x + 3\), \(g(f(x))=10^{x}+3\), which is not a straight line in semi - log plot (since \(\log(10^{x}+3)\) is not linear in \(x\)). Option B: \(g(x)=3\cdot10^{x}\), \(g(f(x))=3\cdot10^{10^{x}}\), which is a very fast - growing function, not a straight line. Option C: \(g(x)=\frac{1}{3}x\), \(g(f(x))=\frac{1}{3}\times10^{x}\), \(\log(y)=\log(\frac{1}{3})+x\), but when \(x = 0\), \(y=\frac{1}{3} eq3\), so it doesn't match the point \((0,3)\). Option D: \(g(x)=3x\), \(g(f(x)) = 3\times10^{x}\), when \(x = 0\), \(y = 3\times1=3\) (wait, no, \(f(0)=1\), \(g(1)=3\times1 = 3\)), which matches the point \((0,3)\) on the graph. So the correct option is D.

Answer:

D. \(g(x)=3x\)